CLASSICAL ELECTRODYNAMICS
for Undergraduates
Professor John W. Norbury
Physics Department
University of Wisconsin-Milwaukee
P.O. Box 413
Milwaukee, WI 53201
1997
Contents
1 MATRICES
1.1 Einstein Summation Convention
1.2 Coupled Equations and Matrices
1.3 Determinants and Inverse . . . .
1.4 Solution of Coupled Equations .
1.5 Summary . . . . . . . . . . . . .
1.6 Problems . . . . . . . . . . . . .
1.7 Answers . . . . . . . . . . . . . .
1.8 Solutions . . . . . . . . . . . . .
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5
5
6
8
11
11
13
14
15
2 VECTORS
19
2.1 Basis Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.2 Scalar Product . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.3 Vector Product . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.4 Triple and Mixed Products . . . . . . . . . . . . . . . . . . . 25
2.5 Div, Grad and Curl (differential calculus for vectors) . . . . . 26
2.6 Integrals of Div, Grad and Curl . . . . . . . . . . . . . . . . . 31
2.6.1 Fundamental Theorem of Gradients . . . . . . . . . . 32
2.6.2 Gauss’ theorem (Fundamental theorem of Divergence) 34
2.6.3 Stokes’ theorem (Fundamental theorem of curl) . . . . 35
2.7 Potential Theory . . . . . . . . . . . . . . . . . . . . . . . . . 36
2.8 Curvilinear Coordinates . . . . . . . . . . . . . . . . . . . . . 37
2.8.1 Plane Cartesian (Rectangular) Coordinates . . . . . . 37
2.8.2 Three dimensional Cartesian Coordinates . . . . . . . 38
2.8.3 Plane (2-dimensional) Polar Coordinates . . . . . . . 38
2.8.4 Spherical (3-dimensional) Polar Coordinates . . . . . 40
2.8.5 Cylindrical (3-dimensional) Polar Coordinates . . . . 41
2.8.6 Div, Grad and Curl in Curvilinear Coordinates . . . . 43
1
2
CONTENTS
2.9
2.10
2.11
2.12
2.13
Summary . . . . . . . . . . .
Problems . . . . . . . . . . .
Answers . . . . . . . . . . . .
Solutions . . . . . . . . . . .
Figure captions for chapter 2
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3 MAXWELL’S EQUATIONS
3.1 Maxwell’s equations in differential form
3.2 Maxwell’s equations in integral form . .
3.3 Charge Conservation . . . . . . . . . . .
3.4 Electromagnetic Waves . . . . . . . . .
3.5 Scalar and Vector Potential . . . . . . .
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43
44
46
47
51
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53
54
56
57
58
60
4 ELECTROSTATICS
4.1 Equations for electrostatics . . . . . . . . . . . . . .
4.2 Electric Field . . . . . . . . . . . . . . . . . . . . . .
4.3 Electric Scalar potential . . . . . . . . . . . . . . . .
4.4 Potential Energy . . . . . . . . . . . . . . . . . . . .
4.4.1 Arbitrariness of zero point of potential energy
4.4.2 Work done in assembling a system of charges
4.5 Multipole Expansion . . . . . . . . . . . . . . . . . .
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63
63
66
68
70
74
74
76
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5 Magnetostatics
5.1 Equation for Magnetostatics . . . . . . . .
5.1.1 Equations from Ampèré’s Law . .
5.1.2 Equations from Gauss’ Law . . . .
5.2 Magnetic Field from the Biot-Savart Law
5.3 Magnetic Field from Ampèré’s Law . . . .
5.4 Magnetic Field from Vector Potential . . .
5.5 Units . . . . . . . . . . . . . . . . . . . . .
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77
77
78
78
79
81
81
81
6 ELECTRO- AND MAGNETOSTATICS IN MATTER
6.1 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Maxwell’s Equations in Matter . . . . . . . . . . . . . . .
6.2.1 Electrostatics . . . . . . . . . . . . . . . . . . . . .
6.2.2 Magnetostatics . . . . . . . . . . . . . . . . . . . .
6.2.3 Summary of Maxwell’s Equations . . . . . . . . . .
6.3 Further Dimennsional of Electrostatics . . . . . . . . . . .
6.3.1 Dipoles in Electric Field . . . . . . . . . . . . . . .
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83
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85
86
88
89
89
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CONTENTS
6.3.2
6.3.3
3
Energy Stored in a Dielectric . . . . . . . . . . . . . .
Potential of a Polarized Dielectric . . . . . . . . . . .
90
91
7 ELECTRODYNAMICS AND MAGNETODYNAMICS
93
7.0.4 Faradays’s Law of Induction . . . . . . . . . . . . . . . 93
7.0.5 Analogy between Faraday field and Magnetostatics . . 96
7.1 Ohm’s Law and Electrostatic Force . . . . . . . . . . . . . . . 97
8 MAGNETOSTATICS
101
9 ELECTRO- & MAGNETOSTATICS IN MATTER
103
10 ELECTRODYNAMICS AND MAGNETODYNAMICS
105
11 ELECTROMAGNETIC WAVES
107
12 SPECIAL RELATIVITY
109
4
CONTENTS
Chapter 1
MATRICES
1.1
Einstein Summation Convention
Even though we shall not study vectors until chapter 2, we will introduce
simple vectors now so that we can more easily understand the Einstein summation convention.
We are often used to writing vectors in terms of unit basis vectors as
A = Ax î + Ay ĵ + Az k̂.
(1.1)
(see Figs. 2.7 and 2.8.) However we will find it much more convenient instead
to write this as
A = A1 ê1 + A2 ê2 + A3 ê3
(1.2)
where our components (Ax , Ay , Az ) are re-written as (A1 , A2 , A3 ) and the
basis vectors (î, ĵ, k̂) become (ê1 , ê2 , ê3 ). This is more natural when considering other dimensions. For instance in 2 dimensions we would write
A = A1 ê1 + A2 ê2 and in 5 dimensions we would write A = A1 ê1 + A2 ê2 +
A3 ê3 + A4 ê4 + A5 ê5 .
However, even this gets a little clumsy. For example in 10 dimensions we
would have to write out 10 terms. It is much easier to write
A=
N
X
Ai êi
(1.3)
i
where N is the number of dimensions. Notice in this formula that the index
i occurs twice in the expression Ai êi . Einstein noticed this always occurred
and so whenever an index was repeated twice he simply didn’t bother to
5
6
CHAPTER 1. MATRICES
P
write N
was always there for twice repeated
i as well because he just knew itP
indices so that instead of writing A = i Ai êi he would simply write A =
P
Ai êi knowing that there was really a i in the formula, that he wasn’t
bothering to write explcitly. Thus the Einstein summation convention is
defined generally as
Xi Yi ≡
N
X
Xi Yi
(1.4)
i
Let us work out some examples.
————————————————————————————————–
Example 1.1.1 What is Ai Bi in 2 dimensions ?
Solution Ai Bi ≡
P2
i=1 Ai Bi
= A1 B1 + A2 B2
Example 1.1.2 What is Aij Bjk in 3 dimensions ?
Solution We have 3 indices here (i, j, k), but only j is repeated
P
twice and so Aij Bjk ≡ 3j=1 Aij Bjk = Ai1 B1k + Ai2 B2k + Ai3 B3k
————————————————————————————————–
1.2
Coupled Equations and Matrices
Consider the two simultaneous (or coupled) equations
x+y =2
x−y =0
(1.5)
which have the solutions x = 1 and y = 1. A different way of writing these
coupled equations is in terms of objects called matrices,
Ã
1
1
1 −1
!Ã
x
y
!
Ã
=
x+y
x−y
!
Ã
=
2
0
!
(1.6)
Notice how the two matrices on the far left hand side get multiplied together.
The multiplication rule is perhaps clearer if we write
Ã
a b
c d
!Ã
x
y
!
Ã
≡
ax + by
cx + dy
!
(1.7)
1.2. COUPLED EQUATIONS AND MATRICES
7
We have invented these matrices with their rule of ’multpilication’ simply as
a way of writing (1.5) in a fancy form. If we had 3 simultaneous equations
x+y+z =3
x−y+z =1
2x + z = 3
we would write



(1.8)




1 1 1
x
x+y+z
4


 
 

 1 −1 1   y  =  x − y + z  =  2 
2 0 1
z
2x + 0y + z
4
(1.9)
Thus matrix notation is simply a way of writing down simultaneous equations. In the far left hand side of (1.6), (1.7) and (1.9) we have a square
matrix multiplying a column matrix. Equation (1.6) could also be written
as
[A] [X] = [B]
(1.10)
with
Ã
A11 A12
A21 A22
!Ã
x1
x2
!
Ã
≡
A11 x1 + A12 x2
A21 x2 + A22 x2
!
Ã
≡
B1
B2
!
(1.11)
or B1 = A11 x1 + A12 x2 and B2 = A21 x1 + A22 x2 . A shorthand for this is
Bi = Aik xk
(1.12)
which is just a shorthand way of writing matrix multiplication form. Note
xk has 1Ã index
! and is a vector. Thus vectors are often written x = xî + y ĵ
x
. This is the matrix way of writing a vector. (do Problem
or just
y
1.1)
Sometimes we want to multiply two square matrices together. The rule
for doing this is
Ã
Ã
≡
A11 A12
A21 A22
A11 B11 + A12 B22 A11 B12 + A12 B22
A21 B11 + A22 B21 A21 B12 + A22 B22
!Ã
!
Ã
≡
B11 B12
B21 B22
C11 C12
C21 C22
!
!
(1.13)
8
CHAPTER 1. MATRICES
Thus, for example, C11 = A11 B11 + A12 B22 and C21 = A21 B11 + A22 B21
which can be written in shorthand as
Cij = Aik Bkj
(1.14)
which is the matrix multiplication formula for square matrices. This is very
easy to understand as it is just a generalization of (1.12) with an extra index
j tacked on. (do Problems 1.2 and 1.3)
————————————————————————————————–
Example 1.2.1 Show that equation (1.14) gives the correct form
for C21 .
Solution Cij = Aik Bkj . Thus C21 = A2k Bk1 = A21 B11 +
A22 B21 .
Example 1.2.2 Show that Cij = Aik Bjk is the wrong formula
for matrix multiplication.
Solution Let’s work it out for C21 :
C21 = A2k B1k = A21 B11 +A22 B12 . Comparing to the expressions
above we can see that the second term is wrong here.
————————————————————————————————–
1.3
Determinants and Inverse
We now need to discuss matrix determinant and matrix inverse. The determinant for a 2 × 2 matrix is denoted
¯
¯ A
¯ 11
¯
¯ A21
A12
A22
¯
¯
¯
¯ ≡ A11 A22 − A21 A12
¯
(1.15)
and for a 3 × 3 matrix it is
¯
¯ A
¯ 11
¯
¯ A21
¯
¯ A31
A12 A13
A22 A23
A32 A33
¯
¯
¯
¯
¯ ≡ A11 A22 A33 + A12 A23 A31 + A13 A21 A32
¯
¯
−A31 A22 A13 − A21 A12 A33 − A11 A32 A23
(1.16)
1.3. DETERMINANTS AND INVERSE
Ã
9

!
1
1 0

The identity matrix [I] is
for 2 × 2 matrices or  0
0 1
0
3 × 3 matrices etc. I is defined to have the true property of
namely
IB ≡ BI ≡ B
0
1
0
an

0

0  for
1
identity,
(1.17)
where B is any matrix. Exercise: Check this is true by multuplying any
2 × 2 matrix times I.
The inverse of a matrix A is denoted as A−1 and defined such that
AA−1 = A−1 A = I.
(1.18)
The inverse is actually
calculated using

 objects called cofactors [3]. Consider
A11 A12 A13


the matrix A =  A21 A22 A23  . The cofactor of the matrix element
A31 A32 A33
A21 for example is defined as
¯
¯ A
12
¯
¯ A32
2+1 ¯
cof (A21 ) ≡ (−)
A13
A33
¯
¯
¯
¯ = −(A12 A33 − A32 A13 )
¯
(1.19)
The way to get the matrix elements appearing in this determinant is just by
crossing out the rows and columns in which A21 appears in matrix A and
the elements left over go into the cofactor.
————————————————————————————————–
Example 1.3.1 What is the cofactor of A22 ?
Solution
¯
¯ A
11
¯
¯ A31
2+2 ¯
cof (A22 ) ≡ (−)
A13
A33
¯
¯
¯
¯ = A11 A33 − A31 A13
¯
————————————————————————————————–
Finally we get to the matrix inverse. The matrix elements of the inverse
matrix are given by [3]
(A−1 )ij ≡
1
|A| cof (Aji )
10
CHAPTER 1. MATRICES
(1.20)
Notice that the ij matrix element of the inverse is given by the ji cofactor.
————————————————————————————————–
Ã
Example 1.3.2 Find the inverse of
answer.
Ã
Solution Let’s write A =
A11 A12
A21 A22
1
1
1 −1
!
Ã
=
!
and check your
1
1
1 −1
!
.
Now cof (A11 ) ≡ (−)1+1 |A22 | = A22 = −1. Notice that the
determinant in the cofactor is just a single number for 2 × 2
matrices. The other cofactors are
cof (A12 ) ≡ (−)1+2 |A21 | = −A21 = −1
cof (A21 ) ≡ (−)2+1 |A12 | = −A12 = −1
cof (A22 ) ≡ (−)2+2 |A11 | = A11 = 1 .
The determinant of A is |A| = A11 A22 − A21 A12 = −1 − 1 = −2.
Thus from (1.20)
(A−1 )11 ≡
(A−1 )12 ≡
(A−1 )21 ≡
(A−1 )22
≡
1
−2 cof (A11 )
1
−2 cof (A21 )
1
−2 cof (A12 )
1
−2 cof (A22 )
Ã
Thus A−1 =
1
2
=
1
2
1
2
1
2
=
− 12
=
=
1
1
1 −1
,
,
,
!
.
.
We can check our answer by making sure that AA−1 = I as
follows,
Ã
AA−1
Ã
=
1
2
=
2 0
0 2
1
1
1 −1
!
Ã
=
!
Ã
1
2
1 0
0 1
1
1
1 −1
!
!
Ã
=
1
2
1
1
1 −1
!Ã
1
1
1 −1
!
.
Thus we are now sure that our calculation of A−1 is correct.
Also having verified this gives justification for believing in all the
formulas for cofactors and inverse given above. (do Problems
1.4 and 1.5)
————————————————————————————————–
1.4. SOLUTION OF COUPLED EQUATIONS
1.4
11
Solution of Coupled Equations
By now we have developed quite a few skills with matrices, but what is the
point of it all ? Well it allows us to solve simultaneous (coupled) equations
(such Ãas (1.5)) !with matrix
(1.6)) as AX = Y where
à methods.
!
à Writing
!
1
1
x
2
A =
,X =
,Y =
, we see that the solution we
1 −1
y
0
want is the value for x and y. In other words we want to find the column
matrix X. We have
AX = Y
(1.21)
so that
Thus
A−1 AX = A−1 Y.
(1.22)
X = A−1 Y
(1.23)
where we have used (1.18).
————————————————————————————————–
Example 1.4.1 Solve the set of coupled equations (1.5) with
matrix methods.
Solution
is re-written
in
à Equation
! (1.5) Ã
!
à (1.6)
! as AX = Y with
1
1
x
2
A =
,X =
,Y =
. We want X =
1 −1
y
0
Ã
A−1 Y
. From Example 1.3.2 we have
X = A−1 Y means
!
Ã
!Ã
!
x
1
1
2
= 12
=
y
1 −1
0
Ã
Ã
1
2
A−1
2
2
=
!
1
2
Ã
=
1
1
1
1
1 −1
!
. Thus
!
.
Thus x = 1 and y = 1.
————————————————————————————————–
1.5
Summary
This concludes our discussion of matrices. Just to summarize a little, firstly
it is more than reasonable to simply think of matrices as another way of
writing and solving simultaneous equations. Secondly, our invention of how
to multiply matrices in (1.12) was invented only so that it reproduces the
12
CHAPTER 1. MATRICES
coupled equations (1.5). For multiplying two square matrices equation (1.14)
is just (1.12) with another index tacked on. Thirdly, even though we did not
prove mathematically that the inverse is given by (1.20), nevertheless we can
believe in the formula becasue we always found that using it gives AA−1 = I.
It doesn’t matter how you get A−1 , as long as AA−1 = I you know that you
have found the right answer. A proof of (1.20) can be found in mathematics
books [3, 4].
1.6. PROBLEMS
1.6
13
Problems
1.1 Show that Bi = Aik xk gives (1.11).
1.2 Show that Cij = Aik Bkj gives (1.13).
1.3 Show that matrix multiplication is non-commutative, i.e. AB 6= BA.
Ã
1.4 Find the inverse of

1 1
0 1
!
and check your answer.

1 1 1


1.5 Find the inverse of  1 −1 1  and check your answer.
2 0 1
1.6 Solve the following simultaneous equations with matrix methods:
x+y+z =4
x−y+z =2
2x + z = 4
14
CHAPTER 1. MATRICES
1.7
Ã
1.4


1.5 
Answers
1 −1
0
1
− 12
1
2
1
!
− 12
− 12
1

1

0 
−1
1.6 x = 1, y = 1, z = 2.
1.8. SOLUTIONS
1.8
15
Solutions
1.1
Bi = Aik xk . We simply evaluate each term. Thus
B1 = A1k xk = A11 x1 + A12 x2
B2 = A2k xk = A21 x1 + A22 x2 .
1.2
Cij = Aik Bkj .
C11 = A1k Bk1
C12 = A1k Bk2
C21 = A2k Bk1
C22 = A2k Bk2
Again just evaluate each term. Thus
= A11 B11 + A12 B21
= A11 B12 + A12 B22
= A21 B11 + A22 B21
= A21 B12 + A22 B22 .
1.3
This can!be
Ã
à seen by
! just
à multiplying
! any two matrices, say
1 2
3 4
Ã
5 6
7 8
!Ã
=
19 22
33 50
!
Ã
.
!
5 6
1 2
23 34
=
.
Whereas
7 8
3 4
31 46
showing that matrix multiplication does not commute.
16
CHAPTER 1. MATRICES
1.4
Ã
!
Ã
!
A11 A12
1 1
Let’s write A =
=
. Now cof (A11 ) ≡
0 1
A21 A22
(−)1+1 |A22 | = A22 = 1. Notice that the determinant in the
cofactor is just a single number for 2 × 2 matrices. The other
cofactors are
cof (A12 ) ≡ (−)1+2 |A21 | = −A21 = 0
cof (A21 ) ≡ (−)2+1 |A12 | = −A12 = −1
cof (A22 ) ≡ (−)2+2 |A11 | = A11 = 1 .
The determinant of A is |A| = A11 A22 − A21 A12 = 1. Thus from
(1.20)
(A−1 )11 ≡ 11 cof (A11 ) = 1 ,
(A−1 )12 ≡ 11 cof (A21 ) = −1 ,
(A−1 )21 ≡ 11 cof (A12 ) = 0 ,
(A−1 )22 ≡ 11 cof (A22 ) = 1 .
Ã
Thus A−1 =
1 −1
0
1
!
.
We can check our answer by making sure that AA−1 = I as
follows,
Ã
!Ã
!
Ã
!
1 1
1 −1
1 0
=
=
= I. Thus we are
0 1
0
1
0 1
now sure that our answer for A−1 is correct.
AA−1
1.8. SOLUTIONS
1.5
17




A11 A12 A13
1 1 1

 

Let’s write A =  A21 A22 A23  =  1 −1 1  .
2 0 1
A31 A32 A33
The cofactors are
¯
¯ A
¯
cof (A11 ) = (−)1+1 ¯ 22
¯ A32
A23
A33
¯
¯
¯
¯ = +(A22 A33 − A32 A23 ) = −1 − 0 = −1
¯
¯
¯ A
21
¯
¯ A31
A23
A33
¯
¯
¯
¯ = −(A21 A33 − A31 A23 ) = −1 + 2 = 1
¯
¯
¯ A
¯
cof (A13 ) = (−)1+3 ¯ 21
¯ A31
A22
A32
¯
¯
¯
¯ = +(A21 A32 − A31 A22 ) = 0 + 2 = 2
¯
¯
¯ A
12
¯
¯ A32
A13
A33
¯
¯
¯
¯ = −(A12 A33 − A32 A13 ) = −1 + 0 = −1
¯
¯
¯ A
11
¯
¯ A31
A13
A33
¯
¯
¯
¯ = +(A11 A33 − A31 A13 ) = 1 − 2 = −1
¯
¯
¯ A
¯
cof (A23 ) = (−)2+3 ¯ 11
¯ A31
A12
A32
¯
¯
¯
¯ = −(A11 A32 − A31 A12 ) = −0 + 2 = 2
¯
¯
¯ A
12
¯
¯ A22
A13
A23
¯
¯
¯
¯ = +(A12 A23 − A22 A13 ) = 1 + 1 = 2
¯
¯
¯ A
¯
cof (A32 ) = (−)3+2 ¯ 11
¯ A21
A13
A23
¯
¯
¯
¯ = −(A11 A23 − A21 A13 ) = −1 + 1 = 0
¯
cof (A12 ) =
¯
(−)1+2
cof (A21 ) =
¯
(−)2+1
cof (A22 ) =
¯
(−)2+2
cof (A31 ) =
¯
(−)3+1
¯
¯ A
11
¯
¯ A21
¯
A12 ¯¯
cof (A33 ) =
¯ = +(A11 A22 − A21 A12 ) = −1 − 1 = −2
A22 ¯
The determinant of A is (see equation (1.16)) |A| = 2. Thus from (1.20)
(A−1 )11 ≡ 12 cof (A11 ) = − 12 ,
(A−1 )12 ≡ 12 cof (A21 ) = − 12 ,
(A−1 )13 ≡ 12 cof (A31 ) = 1 ,
¯
(−)3+3
18
CHAPTER 1. MATRICES
(A−1 )21
(A−1 )22
(A−1 )23
(A−1 )31
(A−1 )32
(A−1 )33
≡
≡
≡
≡
≡
≡
1
1
2 cof (A12 ) = 2 ,
1
1
2 cof (A22 ) = − 2
1
2 cof (A32 ) = 0 ,
1
2 cof (A13 ) = 1 ,
1
2 cof (A23 ) = 1 .
1
(A33 ) = −1
2 cof

− 1 − 12
 1 2
−1
− 12
A = 2
.
, 
1

0 .
Thus
1
1 −1
−1 = I as follows,
We can check
our
answer
that AA

 by making sure 


1 1 1
− 12 − 12
1
1 0 0





− 12
0  =  0 1 0  = I. Thus
AA−1 =  1 −1 1   12
2 0 1
0 0 1
1
1 −1
we are now sure that our answer for A−1 is correct.
1.6
AX = Y is written

  as

1

 1
2
Thus


1 1
x
4

 

−1 1   y  =  2 
0 1
z
4
−1
−1
X
 =A
 Yand we found A in problem
 
1
1
x
−2 −2
1
4

 

 
− 12
0  2  = 
Thus  y  =  12
z
4
1
1 −1
Thus the solution is x = 1, y = 1, z = 2.
1.5.

1

1 .
2
Chapter 2
VECTORS
In this review chapter we shall assume some familiarity with vectors from
fresman physics. Extra details can be found in the references [1].
2.1
Basis Vectors
We shall assume that we can always write a vector in terms of a set of basis
vectors
A = Ax î + Ay ĵ + Az k̂ = A1 ê1 + A2 ê2 + A3 ê3 = Ai êi
(2.1)
where the components of the vector A are (Ax , Ay , Az ) or (A1 , A2 , A3 ) and
the basis vectors are (î, ĵ, k̂) or (ê1 , ê2 , ê3 ). The index notation Ai and êi is
preferrred because it is easy to handle any number of dimensions.
In equation (2.1) we are using the Einstein summation convention for
repeated indices which says that
xi yi ≡
X
xi yi .
(2.2)
i
In other words when indices are repeated it means that a sum is always
implied. We shall use this convention throughout this book.
With basis vectors it is always easy to add and subtract vectors
A + B = Ai êi + Bi êi = (Ai + Bi )êi .
(2.3)
Thus the components of A + B are obtained by adding the components of
A and B separately. Similarly, for example,
A − 2B = Ai êi − 2Bi êi = (Ai − 2Bi )êi .
19
(2.4)
20
2.2
CHAPTER 2. VECTORS
Scalar Product
Let us define the scalar product (also often called the inner product) of two
vectors as
A.B ≡ AB cos θ
(2.5)
where A ≡ |A| is the magnitude of A and θ is the angle between A and B.
(Here || means magnitude and not deteminant). Thus
A.B = Ai êi .Bj êj
(2.6)
Note that when we are multiplying two quantities that both use the Einstein
summation convention, we must use different indices. (We were OK before
when adding). To see this let’s work out (2.6) explicitly in two dimensions.
A = A1 ê1 +A2 ê2 and B = B1 ê1 +B2 ê2 so that A.B = (A1 ê1 +A2 ê2 ).(B1 ê1 +
B2 ê2 ) = A1 B1 ê1 .ê1 + A1 B2 ê1 .ê2 + A2 B1 ê2 .ê1 + A2 B2 ê2 .ê2 which is exactly
what you get when expanding out (2.6). However if we had mistakenly
written A.B = Ai êi .Bi êi = Ai Bi êi .êi = A1 B1 ê1 .ê1 + A2 B2 ê2 .ê2 which is
wrong. A basic rule of thumb is that it’s OK to have double repeated indices
but it’s never OK to have more.
Let’s return to our scalar product
A.B = Ai êi .Bj êj
= (êi .êj )Ai Bj
≡ gij Ai Bj
(2.7)
where we define a quantitiy gij called the metric tensor as gij ≡ êi .êj . Note
that vector components Ai have one index, scalars never have any indices
and matrix elements have two indices Aij . Thus scalars are called tensors
of rank zero, vectors are called tensors of rank one and some matrices are
tensors of rank two. Not all matrices are tensors because they must also
satisfy the tensor transformation rules [1] which we will not go into here.
However all tensors of rank two can be written as matrices. There are also
tensors of rank three Aijk etc. Tensors of rank three are called tensors of
rank three. They do not have a special name like scalar and vector. The
same is true for tensors of rank higher than three.
2.2. SCALAR PRODUCT
21
Now if we choose our basis vectors êi to be of unit length |êi | = 1 and
orthogonal to each other then by (2.5)
êi .êj = |êi ||êj | cos θ = cos θ = δij
(2.8)
where δij is defined as
δij ≡ 1
if
i=j
≡0
if
i 6= j
which can also be written as a matrix
Ã
[δij ] =
1 0
0 1
(2.9)
!
(2.10)
for two dimensions. Thus if gij = δij then we have what is called a Cartesian
space (or flat space), so that
A.B = gij Ai Bj = δij Ai Bj
= δi1 Ai B1 + δi2 Ai B2
= δ11 A1 B1 + δ21 A2 B1 + δ12 A1 B2 + δ22 A2 B2
= A1 B1 + A2 B2 ≡ Ai Bi .
(2.11)
Thus
A.B = Ai Bi
(2.12)
Now A.B = Ai Bi = Ax Bx + Ay By is just the scalar product that we are
used to from freshman physics, and so Pythagoras’ theorem follows as
A.A ≡ A2 = Ai Ai = A2x + A2y .
(2.13)
Note that the fundamental relations of the scalar product (2.12) and the
form of Pythagoras’ theorem
à follow!directly from our specification of the
1 0
metric tensor as gij = δij =
.
0 1
As an aside, note that
easily have defined a non-Cartesian
à we could
!
1 1
in which case Pythagoras’ theorem would
space, for example gij =
0 1
change to
(2.14)
A.A ≡ A2 = Ai Ai = A2x + A2y + Ax Ay .
22
CHAPTER 2. VECTORS
Thus it is the metric tensor gij ≡ êi .êj given by the scalar product of the
unit vectors which (almost) completely defines the vector space that we are
considering.
2.3
Vector Product
In the previous section we ’multiplied’ two vectors to get a scalar A.B. However if we start with two vectors maybe we can also define a ’multiplication’
that results in a vector, which is our only other choice. This is called the
vector product or cross product denoted as A × B. The magnitude of the
vector product is defined as
|A × B| ≡ AB sin θ
(2.15)
whereas the direction of C = A × B is defined as being given by the right
hand rule, whereby you hold the thumb, fore-finger and middle finger of your
right hand all at right angles to each other. The thumb represents vector C,
the fore-finger represents A and the middle finger represents B.
————————————————————————————————–
Example 2.3.1 If D is a vector pointing to the right of the page
and E points down the page, what is the direction of D × E and
E × D?
Solution D is the fore-finger, E is the middle finger and so
D × E which is represented by the thumb ends up pointing into
the page. For E×D we swap fingers and the thumb E×D points
out of the page. (do Problem 2.1)
————————————————————————————————–
From our definitions above for the magnitude and direction of the vector
product it follows that
ê1 × ê1 = ê2 × ê2 = ê3 × ê3 = 0
(2.16)
because θ = 0o for these cases. Also
ê1 × ê2 = ê3 = −ê2 × ê1
ê2 × ê3 = ê1 = −ê3 × ê2
ê3 × ê1 = ê2 = −ê1 × ê3
(2.17)
2.3. VECTOR PRODUCT
23
which follows from the right hand rule and also because θ = 90o .
Let us now introduce some short hand notation in the form of the LeviCivitá symbol (not a tensor ??) defined as
²ijk ≡ +1
if ijk are in the order of 123 (even permutation)
≡ −1
≡0
if ijk not in the order of 123 (odd permutation)
if any of ijk are repeated (not a permutation)(2.18)
For example ²123 = +1 because 123 are in order. Also ²231 = +1 because
the numbers are still in order whereas ²312 = −1 becasue 312 are out of
numerical sequence. ²122 = ²233 = 0 etc., because the numberse are not
a permutation of 123 because two numbers are repeated. Also note that
²ijk = ²jki = ²kij = −²ikj etc. That is we can permute the indices without
changing the answer, but we get a minus sign if we only swap the places of
two indices. (do Problem 2.2)
We can now write down the vector product of our basis vectors as
êi × êj = ²ijk êk
(2.19)
where a sum over k is implied because it is a repeated index.
————————————————————————————————–
Example 2.3.2 Using (2.19) show that ê2 ×ê3 = ê1 and ê2 ×ê1 =
−ê3 and ê1 × ê1 = 0.
Solution From (2.19) we have
ê2 × ê3 = ²23k êk
= ²231 ê1 + ²232 ê2 + ²233 ê3
= +1ê1 + 0ê2 + 0ê3
= ê1 .
ê2 × ê1 = ²21k êk
= ²211 ê1 + ²212 ê2 + ²213 ê3
= 0ê1 + 0ê2 − 1ê3
= −ê3 .
ê1 × ê1 = ²11k êk
= 0.
because ²11k = 0 no matter what the value of k.
24
CHAPTER 2. VECTORS
————————————————————————————————–
Using (2.19) we can now write the vector product as A × B = Ai êi ×
Bj êj = Ai Bj êi × êj = Ai Bj ²ijk êk Thus our general formula for the vector
product is
A × B = ²ijk Ai Bj êk
(2.20)
The advantage of this formula is that it gives both the magnitude and direction. Note that the kth component of A × B is just the coefficient in front
of êk , i.e.
(2.21)
(A × B)k = ²ijk Ai Bj
————————————————————————————————–
Example 2.3.3 Evaluate the x component of (2.20) explicitly.
Solution The right hand side of (2.20) has 3 sets of twice reP P P
P
peated indices ijk which implies i j k . Let’s do k first.
A×B = ²ij1 Ai Bj ê1 +²ij2 Ai Bj ê2 +²ij3 Ai Bj ê3 . The x component
of A × B is just the coefficient in front of ê1 . Let’s do the sum
over i first. Thus
(A × B)1 = ²ij1 Ai Bj
= ²1j1 A1 Bj + ²2j1 A2 Bj + ²3j1 A3 Bj +
= ²111 A1 B1 + ²121 A1 B2 + ²131 A1 B3 +
²211 A2 B1 + ²221 A2 B2 + ²231 A2 B3 +
²311 A3 B1 + ²321 A3 B2 + ²331 A3 B3
= 0A1 B1 + 0A1 B2 + 0A1 B3 +
0A2 B1 + 0A2 B2 + 1A2 B3 +
0A3 B1 − 1A3 B2 + 0A3 B3
= A2 B3 − A3 B2 .
————————————————————————————————–
2.4. TRIPLE AND MIXED PRODUCTS
25
Similarly one can show (do Problem 2.3) that (A × B)2 = A3 B1 − A1 B3
and (A × B)3 = A1 B2 − A2 B1 , or in other words
A × B = (A2 B3 − A3 B2 )ê1 + (A3 B1 − A1 B3 )ê2 + (A1 B2 − A2 B1 )ê3 (2.22)
which is simply the result of working out (2.20). The formula (2.20) can
perhaps be more easily memorized by writing it as a determinant
¯
¯ ê
¯ 1
¯
A × B = ¯ A1
¯
¯ B1
¯
ê2 ê3 ¯¯
¯
A2 A3 ¯
¯
B2 B3 ¯
(2.23)
This doesn’t come from the theory of matrices or anything complicated.
It is just a memory device for getting (2.20). (do Problem 2.4)
2.4
Triple and Mixed Products
We will now consider triples and mixtures of scalar and vector products such
as A.(B × C) and A × (B × C) etc. Our kronecker delta δij and Levi-Civitá
²ijk symbols will make these much easier to work out. We shall simply show
a few examples and you can do some problems. However before proceeding
there is a useful identity that we shall use namely
²kij ²klm = δil δjm − δim δjl .
(2.24)
To show this is true it’s easiest to just work out the left and right hand sides
for special index values.
————————————————————————————————–
Example 2.4.1 Show that (2.24) is true for i = 1, j = 2, l =
3, m = 3.
Solution The left hand side is
²kij ²klm = ²k12 ²k33
= ²112 ²133 + ²212 ²233 + ²312 ²333
= (0)(0) + (0)(0) + (+1)(0)
= 0.
(2.25)
The right hand side is
δil δjm − δim δjl = δ13 δ23 − δ13 δ23
= (0)(0) − (0)(0)
= 0.
(2.26)
26
CHAPTER 2. VECTORS
Thus the left hand side equals the right hand side. (do Problem
2.5)
Example 2.4.2 Show that A.(B × C) = B.(C × A).
Solution
A.(B × C) = Ak (B × C)k = Ak ²ijk Bi Cj
= Bi ²ijk Cj Ak
= Bi ²jki Cj Ak
= Bi (C × A)i
= B.(C × A)
(2.27)
where we used the fact that ²ijk = ²jki . (do Problem 2.6)
————————————————————————————————–
2.5
Div, Grad and Curl (differential calculus for
vectors)
Some references for this section are the books by Griffiths and Arfken [2, 8]
(x)
We have learned about the derivative dfdx
of the function f (x) in elementary
calculus. Recall that the derivative gives us the rate of change of the function
in the direction of x. If f is a function of three variables f (x, y, z) we can
∂f ∂f
form partial derivatives ∂f
∂x , ∂y , ∂z which tell us the rate of change of f
in three different diections. Recall that the vector components Ax , Ay , Az
tell us the size of the vector A in three different diections. Thus we expect
that the three partial derivatives may be interpreted as the components of
a vector derivative denoted by 5 and defined by
∂
5 ≡ êi ∂x
≡ êi 5i
i
(2.28)
with 5i ≡
∂
∂xi .
Expanding (with the Einstein summation convention) gives
∂
∂
∂
5=
+ ê2 ∂x∂ 2 + ê3 ∂x∂ 3 = î ∂x
+ ĵ ∂y
+ k̂ ∂z
. (Note that it’s important
∂
to write the ∂xi to the right of the êi , otherwise we might mistakenly think
ê1 ∂x∂ 1
2.5. DIV, GRAD AND CURL (DIFFERENTIAL CALCULUS FOR VECTORS)27
∂
that ∂x
acts on êi .) Even though we use the standard vector symbol for 5
i
it is not quite the same type of vector as A. Rather 5 is a vector derivative
or vector operator. Actually its proper mathematical name is a dual vector
or a one-form.
Now that we are armed with this new vector operator weapon, let’s start
using it. The two types of objects that we know about so far are functions
φ(x, y, z) and vectors A. The only object we can form with 5 acting on φ
is called the gradient of the function φ and is defined by
5φ ≡ êi
∂φ
∂xi
(2.29)
∂φ
∂φ
∂φ
∂φ
∂φ
∂φ
= ê1 ∂x
+ ê2 ∂x
+ ê3 ∂x
= î ∂φ
which is expanded as 5φ ≡ êi ∂x
∂x + ĵ ∂y + k̂ ∂z .
1
2
3
i
Now let’s consider how 5 acts on vectors. The only two choices we have
for vector ’multiplication’ are the scalar product and the vector product.
The scalar product gives the divergence of a vector defined as
5.A ≡ 5i Ai ≡
∂Ai
∂xi
(2.30)
expanded as 5.A ≡ 5i Ai = 51 A1 + 52 A2 + 53 A3 = 5x Ax + 5y Ay + 5z Az
∂Ay
∂A2
∂A3
∂Ax
∂Az
1
or equivalently 5.A = ∂A
∂x1 + ∂x2 + ∂x3 = ∂x + ∂y + ∂z .
The vector product gives the curl of a vector defined as
5 × A ≡ ²ijk (5i Aj )êk = ²ijk
3
expanded as 5 × A = ( ∂A
∂x2 −
∂Ay
∂z )î
∂Aj
êk
∂xi
∂A2
∂A1
∂A3
∂x3 )ê1 + ( ∂x3 − ∂x1 )ê2 +
∂Ay
∂Az
∂Ax
x
( ∂A
∂z − ∂x )ĵ + ( ∂x − ∂y )k̂.
(2.31)
2
( ∂A
∂x1 −
∂A1
∂x2 )ê3
or
5×A=
−
+
Thus the divergence, gradient and curl (often abbreviated as div, grad
and curl) are the only ways of possibly combining 5 with vectors and functions.
Scalar and Vector Fields. In order for equation (2.29) to make any sense
it is obvious that the function must depend on the variables x, y, z as φ =
φ(x, y, z). This type of function is often called a scalar field because the
value of the scalar is different at different points x, y, z. There is nothing
complicated here. The scalar field φ(x, y, z) is just the ordinary function of
three variables that all calculus students know about. However the vector A
that we are talking about in equations (2.30) and (2.31) is not a fixed vector
such as A = 3î + 2ĵ + 4k̂ that we saw in introductory physics. (It could be,
but then 5.A = 5 × A = 0 which is the trivial case. We want to be more
z
( ∂A
∂y
28
CHAPTER 2. VECTORS
general than this.) Rather A is, in general, a vector field A = A(x, y, z)
where each component of A is itself a scalar field such as Ax (x, y, z). Thus
A = A(x, y, z) = Ax (x, y, z)î+Ay (x, y, z)ĵ +Az (x, y, z)k̂. The careful reader
will have already noticed that this must be the case if div and curl are to
be non-zero. Note that whereas B = 3î + 2ĵ + 4k̂ is a same arrow at all
points in space, the vector field A(x, y, z) consists of a different arrow at all
points in space. For example, suppose A(x, y, z) = x2 y î + z 2 ĵ + xyz k̂. Then
A(1, 1, 1) = î + ĵ + k̂, but A(0, 1, 1) = ĵ etc.
————————————————————————————————–
Example 2.5.1 Sketch a representative sample of vectors from
the vector fields A(x, y) = xî+y ĵ, B = k̂ and C(x, y) = −y î+xĵ.
(These examples are form reference [2].)
Solution A(x, y) is evaluated at a variety of points such as
A(0, 0) = 0, A(0, 1) = ĵ, A(1, 0) = î , A(1, 1) = î + ĵ, A(0, 2) =
2ĵetc. We draw the corresponding arrow at each point to give
the result shown in Fig. 2.1.
For the second case B = ĵ the vector is independent of the coordinates (x, y, z). This means that B = ĵ at every point in space
and is illustrated in Fig. 2.2.
Finally for C(x, y) we have C(1, 1) = −î + ĵ, C(1, −1) = −î − ĵ,
C(−1, 1) = î + ĵ etc. This is shown in Fig. 2.3.
————————————————————————————————–
Now let’s study the very important physical implications of div, grad and
curl [2, 5].
Physical interpretation of Gradient. We can most easily understand the
meaning of 5φ by considering the change in the function dφ corresponding
to a change in position dl which is written dl = dxî + dy ĵ + dz k̂ [2, 8]. If
φ = φ(x, y, z) then from elementary calculus it’s change is given by dφ =
∂φ
∂φ
∂φ
∂φ
∂xi dxi = ∂x dx + ∂y dy + ∂z dz, which is nothing more than
dφ = (5φ).dl = |5φ||dl| cos θ
(2.32)
Concerning the direction of 5φ , is can be seen that dφ will be a maximum
when cos θ = 1, i.e. when dl is chosen parallel to 5φ. Thus if I move in
the same direction as the gradient, then φ changes maximally. Therefore
the direction of 5φ is along the greatest increase of φ. (think of surfaces of
constant φ being given by countour lines on a map. Then the direction of the
2.5. DIV, GRAD AND CURL (DIFFERENTIAL CALCULUS FOR VECTORS)29
gradient is where the contour lines are closest together. ) The magnitude
∂φ
|5φ| gives the slope along this direction. (Each component of dφ = ∂x
dxi =
i
∂φ
∂φ
î ∂φ
∂x + ĵ ∂y + k̂ ∂z clearly gives a slope.)
The gradient of a function is most easily visualized for a two dimensional
finction φ(x, y) where x and y are the latitude and longtitude and φ is the
height of a hill. In this case surfaces of constant φ will just be like the contour
lines on a map. Given our discovery above that the direction of the gradient
is in the direction of steepest ascent and the magnitude is the slope in this
direction, then it is obvious that if we let a smooth rock roll down a hill then
it will start to roll in the direction of the gradient with a speed proportional
to the magnitude of the gradient. Thus the direction and magnitude of the
gradient is the same as the direction and speed that a rock will take when
rolling freely down a hill. If the gradient vanishes, then that means that you
are standing on a local flat spot such as a summit or valley and a rock will
remain stationary at that spot.
————————————————————————————————–
Example 2.5.2 Consider the simple scalar field φ(x, y) ≡ x.
Compute the gradient and show that the magnitude and direction are consistent with the statemtns above.
Solution Clearly φ only varies in the x direction and does not
change in the y direction. Thus the direction of maximum increase is expected to be soleyl in the î direction. This agreees
with the computation of the gradient as 5φ = î. Every student knows that the straight line φ = x has a slope of 1 which
agrees with the computation of the magnitude as |5φ| = 1. (do
Problem 2.7)
————————————————————————————————–
Physical Interpretation of Divergence. The divergence of a vector field
represents the amount that the vector field spreads out or diverges. Consider
our three examples of vector fields A(x, y) = xî+y ĵ and B = ĵ and C(x, y) =
−y î + xĵ. From the pictures representing these vector fields (see Figs 2.1,
2.2 and 2.3) we can see that A does spread out but B and C do not spread.
Thus we expect that B and C have zero divergence. This is verified by
calculating the divergence explicitly.
————————————————————————————————–
Example 2.5.3 Calculate the divergence of A(x, y) = xî + y ĵ
30
CHAPTER 2. VECTORS
Solution 5.A ≡ 5i Ai =
∂Ax
∂x
+
∂Ay
∂y
z
+ ∂A
∂z =
∂y
∂x
∂x + ∂y
= 1+1 = 2
Example 2.5.4 Calculate the diverbence of B = k̂.
∂B
y
∂Bz
∂
x
Solution 5.B = ∂B
∂x + ∂y + ∂z = 0 + 0 + ∂z (1) = 0 + 0 + 0 = 0
implying that the vector field does not spread out. do Problem
2.8
————————————————————————————————–
Physical Interpretaion of Curl The curl measure the amount of rotation
in a vector field. This could actually be measured by putting a little paddlewheel (with a fixed rotation axis) that is free to rotate in the vector field
(like a little paddlewheel in a swirling water vortex). If the paddlewheel
spins, then the vector field has a non-zero curl, but if the paddlewheel does
not spin, then the vector field has zero curl. From Figs 2.1, 2.2 and 2.3 one
expects that A and B would produce no rotation, but C probably would.
This is verified in the following examples.
————————————————————————————————–
Example 2.5.5 Calculate the curl of A = xî + y ĵ.
∂Ay
∂Ay
∂Ax
∂Az
∂Ax
∂z )î + ( ∂z − ∂x )ĵ + ( ∂x − ∂y )k̂
∂y ∂x
∂x
(0− ∂y
∂z )î+( ∂z −0)ĵ+( ∂x − ∂y )k̂ = (0−0)î+(0−0)ĵ+(0−0)k̂ = 0
z
Solution 5 × A = ( ∂A
∂y −
=
Example 2.5.6 Calculate the curl of C = −y î + xĵ.
∂Cy
∂Cy
∂Cx
∂Cz
∂Cx
∂z )î + ( ∂z − ∂x )ĵ + ( ∂x − ∂y )k̂
∂y
∂x ∂y
(0− ∂x
∂z )î+(− ∂z −0)ĵ+( ∂x + ∂y )k̂ = (0−0)î+(0−0)ĵ+(0−0)k̂ =
z
Solution 5 × C = ( ∂C
∂y −
=
2k̂.
Notice that the direction of the curl (k̂ in this case) is perpendicular to the plane of the vector field C, as befits a vector product.
(do Problem 2.9).
————————————————————————————————–
Second Derivatives. We have so far encountered the scalar 5.A and the
vectors 5φ and 5 × A . We can operate again with 5. However we can
only form the gradient of the scalar (5.A) as 1) 5(5.A), but we can form
the divergence and curl of both the vectors 5φ and 5 × A as 2) 5(5φ) and
3) 5 × (5φ) and 4) 5.(5 × A) and 5) 5 × (5 × A) . However 3) and 4)
are identically zero and 5) contains 1) and 2) already within it. (See [2] for
discussions of this.)
2.6. INTEGRALS OF DIV, GRAD AND CURL
31
The only second derivative that we shall use is 5(5φ) ≡ 52 φ often
called the Laplacian of φ, given by
52 φ =
∂2φ ∂2φ ∂2φ
+ 2 + 2.
∂x2
∂y
∂z
(2.33)
Exercise: Verify that 5 × (5φ) = 0 and 5.(5 × A) = 0.
2.6
Integrals of Div, Grad and Curl
In calculus courses [5] students
will
have learned
about line, surface and
R
R
R
volume integrals such as A.dl or B.dA or f dτ where dl is an oriented
line segment, dA is an oriented area element (which by definition always
points perpendicular to the surface area) and dτ is a volume element. We
shall be interested in obtaining line, surface and volume integrals of div,
grad and curl. One of the main uses of such integrals will be conversion of
Maxwell’s equations from differential to integral form.
In 3-dimensions we have 3 infinitessimal increments of length namely dx,
dy and dz. It is therefore very natural to put them together to form an
infinitessimal length vector as
dl ≡ dlx î + dly ĵ + dlz k̂
= dxî + dy ĵ + dz k̂
(2.34)
It is also natural to combine them all into an infinitessimal volume element
dτ ≡ dxdydz
(2.35)
which is a scalar. How should we form an area element ? Well we could
either use dxdy or dxdz or dydz. These 3 choices again suggest to us to form
a vector
dA ≡ dAx î + dAy ĵ + dAz k̂
= dydz î + dxdz ĵ + dxdy k̂
(2.36)
where
dAx ≡ dydz
dAy ≡ dxdz
dAz ≡ dxdy
(2.37)
32
CHAPTER 2. VECTORS
is the natural choice we would make. Note that, for example, dydz forms an
area in the yz plane but points in the î direction. Thus area is a vector that
points in the direction perpendicular to the surface. Notice that if we had
say 4 dimensions then we could form a 4-vector out of volume elements that
would also have a direction in the 4-dimensional hyperspace.
We will be discussing four important results in this section namely the
fundamental theorem of calculus, the fundamental theorem of gradients, the
fundamental theorem of divergence (also called Gauss’ theorem) and the
fundamental theorem of curl (also called Stokes’ theorem). However we
will leave the proofs of these theorems to mathematics courses. Neverthless
we hope to make these theorems eminently believable via discussion and
examples.
We proceed via analogy with the fundamental theorem of calculus which
states
Z b
df
dx = f (b) − f (a)
(2.38)
a dx
df
has been ’cancelled’ out by the integral over dx to
where the derivative dx
give a right hand side that only depends on the end points of integration .
The vector derivatives we have are 5f , 5.C and 5 × C and we wish to
’cancel’ out the derivatives with an integral.
2.6.1
Fundamental Theorem of Gradients
∂f
∂f
The easiest to deal with is the gradient 5f ≡ î ∂f
∂x + ĵ ∂y + k̂ ∂z because
∂f ∂f
its three components ( ∂f
∂x , ∂y , ∂z ) are just ordinary (partial) derivatives like
that appearing in the left hand side of the fundamental theorem of calculus
equation (2.38). However, because 5f is a vector we can’t just integrate
∂f
over dx as in (2.38). We want to integrate ∂f
∂x over dx and ∂y over dy and
∂f
∂z
over dz and then we will have a three dimensional version of (2.38). The
simplest way to do this is to integrate over dl ≡ îdx + ĵdy + k̂dz to give
∂f
∂f
(5f ).dl = ∂f
∂x dx + ∂y dy + ∂z dz which is a three dimensional version of the
integrand in (2.38). Griffiths [2] calls the result the fundamental theorem of
gradients
Z
(5f ).dl = f (b) − f (a)
(2.39)
which is a result easily understood in light of (2.38). A point ot note is
that f = f (x) in (2.38), but f = f (x, y, z) in (2.39). Thus a and b in
(2.39) actually represent a triple of coordinates. Note again that the right
2.6. INTEGRALS OF DIV, GRAD AND CURL
33
hand sideR of (2.39) depends only on the endpoints. ThreeR dimensional line
integrals dl are different from one dimensional integrals dx in that three
dimensional line integrals with the same end points can be performed over
different paths as shown in Fig. 2.4, whereas the one dimensional integral
always goes straight along the x axis.
Because the right hand side of (2.39) depends only on the end points of
integration, there are two important corollaries of the fundamental theorem
of gradients. The first is that
Rb
a (5f ).dl
is independent of path of integration
(2.40)
and secondly
I
(5f ).dl = 0
(2.41)
H
where dl means that the integral has been performed around a closed loop
with identical endpoints a = b. These
two results follow automatically from
R
our discussions above. Note that (5f ).dl is independent
of the path but
R
this is not true for arbitrary vectors. In general C.dl is not independent of
path [2] (pg.31).
————————————————————————————————–
ExampleR 2.6.1 Let f = xy and a = (0, 0, 0) and b = (1, 1, 0).
Evaluate ba (5f ).dl along two different integration paths and
show that the results are the same.
Solution Let us choose the paths shown in Fig 2.5. First evaluate the integral along the AB path.
R
Path A ( dy = dz = 0): A (5f ).dl =
because y = 0 along this path.
R
Path B ( dx = dz = 0): B (5f ).dl =
1 because x = 1 along this path.
Thus
R
AB (5f ).dl
=
R
A (5f ).dl
Path C ( dz = 0, y = x):
R
R
R1
R
+
R
R 1 ∂f
0 ∂x dx
R 1 ∂f
0 ∂y dy
B (5f ).dl
C (5f ).dl
=
=
=
R1
0 ydx
R1
0 xdy
=
= 0
R1
0 dy
=
= 0 + 1 = 1.
R
( ∂f
∂x dx +
∂f
∂y dy)
=
ydx + xdy = 2 0 xdx = 1 because y = x. This answer is the
same for the AB path. (do Problem 2.10).
34
CHAPTER 2. VECTORS
Example 2.6.2 Check the fundamental theorem of gradients
using the above example.
Solution The above example has evaluated the left hand side of
(2.39). To check the theorem we need to evaluate the right hand
side for f = xy. Thus
f (a) = f (0, 0) = 0
f (b) = f (1, 1) = 1
f (b) − f (a) = 1 which agrees with the calculation in the previous
example. (do Problem 2.11).
————————————————————————————————–
2.6.2
Gauss’ theorem (Fundamental theorem of Divergence)
The divergence 5.C is a scalar quantity, so it is natural to expect that we
integrate it withR a scalar. Out integral elements are dl, dA and dτ , so we
will be wanting (5.C)dτ . The fundamental theorem of divergence [2] (often
also called Gauss’ divergence theorem) is
Z
I
(5.C)dτ =
H
C.dA ≡ Φ0
(2.42)
H
where dA denotes an integral over a closed surface and Φ0 ≡ C.dA is
calledRthe flux. Actually it is the flux over a closed surface. We shall denote
Φ ≡ dA as the ordinary flux. It is easy to understand (2.42) in light
of the fundamental theorem of gradients. Firstly dτ can be thought of as
∂
dτ = dxdydz and so it will ’cancel’ only say ∂x
in 5.C and we will be left
with dydz which is the dA integral on the right hand side of (2.42). We were
∂
unable to get rid of the entire dτ integral because 5 has only things like ∂x
in it, which can only at most convert dτH into dA. Secondly the fact that we
are left with a closed surface integral dA is exactly analagous to the line
integral on the left hand side of (2.39) giving a term on the right hand side
dependent only on the end points . A closed surface encloses a volume just
as two end points enclose a line.
————————————————————————————————–
Example 2.6.3 Check the divergence theorem for C = xî +
yz ĵ +xk̂ using the volume given by the unit cube with one corner
located at the origin.
2.6. INTEGRALS OF DIV, GRAD AND CURL
Solution 5.C =
∂Cx
∂x
+
∂Cy
∂y
∂Cz
∂z
+
= 1 + z + 0 = 1 + z.
The left hand side of (2.42) becomes
R
(5.C).dτ =
R1
R1 R1
0 dx 0 dy 0 dz(1
35
+ z) =
R1
3R 1
2 0 dx 0 dy
= 32 .
To get the right hand side we must evaluate the surface integral
over all six sides of the cube (closed surface). The direction of
each of these
surfaces Ris shown inRFig. 2.6. The
surfaceRintegral is
H
R
R
C.dA
=
C.dA+
C.dA+
C.dA+
C.dA+
A
B
C
D
E C.dA+
R
C.dA
F
For surface A (x = 1): C = î + yz ĵ + k̂, dA = dydz î,
giving
R
A C.dA
=
R1
R1
0 dy 0 dz
= 1.
For surface B (y = 1): C = xî + z ĵ + xk̂, dA = dxdz ĵ,
giving
R
B C.dA
=
R1
R1
0 dx 0 zdz
= 12 .
For surface C (x = 0): C = yz ĵ, dA = −dydz î,
giving
R
C C.dA
= 0.
For surface D (y = 0): C = xî + xk̂, dA = −dxdz ĵ,
giving
R
D C.dA
= 0.
For surface E (z = 0): C = xî + xk̂, dA = −dxdy k̂,
giving
R
E C.dA
R1
=−
R1
0 xdx 0 dy
= − 12 .
For surface F (z = 1): C = xî + y ĵ + xk̂, dA = dxdy k̂,
giving
H
R
F C.dA
R1
1
0 xdx 0 dy = 2 .
1 + 12 + 0 + 0 − 12 + 12
=
R1
Thus C.dA =
=
left hand side. (do Problem 2.12).
3
2
in agreement with the
————————————————————————————————–
2.6.3
Stokes’ theorem (Fundamental theorem of curl)
Finally we integrate the vector 5 × C with our remaining integral element
of area dA. The fundamental theorem of curl (often also called Stokes’
theorem) is
Z
I
(5 × C).dA =
H
C.dl
(2.43)
where dl denotes a line integral over a closed loop. To understand this
∂
result we apply similar reasoning as in the previous section. The ∂x
in 5
36
CHAPTER 2. VECTORS
kills off dx in dA leaving us with dl. Also a closed loop encloses an area just
as a closed area encloses a volume and end points enclose a line. Thus the
right hand sides of (2.43) and (2.42) are analagous to the right hand side of
(2.39).
As with the fundamental theorem of gradients, Stokes’ theorem also has
two corollaries, namely
R
(5 × C).dA is independent of the surface of integration
(2.44)
and
I
(5 × C).dA = 0
(2.45)
which are both analagous to (2.40) and (2.41). One may ask why we didn’t
have two similar corollaries following Gauss’ divergence theorem. The reason
is that the area of (2.44) and (2.45) can be embedded in a volume and the
line of (2.40) and (2.41) can be embedded in an area. Thus we can imagine
different areas and lines to do our integrals. However if we confine ourselves
to our three dimensional world then the volume is as high as we go. It is not
embedded in anything else and so we don’t have different volume integration
paths to choose
from. There is always only one. I suppose we could have
R
said that (5.C)dτ is independent of the volume, but because all volume
paths are identical, it is rather a useless statement.
2.7
Potential Theory
We shall discuss two important theorems from the mathematical subject
known as potential theory [2, 8].
Theorem 1. If E is a vector field, then E = − 5 V iff 5 × E = 0, where
V is a scalar field called the scalar potential.
Theorem 2. If B is a vector field, then B = 5 × A iff 5.B = 0, where
A is a vector field called the vector potential.
(The word iff is shorthand for ’if and only if’.). The minus sign in Theorem 1 is arbitrary. It can simply be left off by writing V = −W where W is
also a scalar potential. The only reason the minus sign appears is because
physicists like to put it there.
So far these theorems have nothing to do with physics. They are just
curious mathematical results at this stage. Also the ’if’ part of the theorems
2.8. CURVILINEAR COORDINATES
37
are easy to prove, but the ’only if’ piece is difficult and won’t be discussed
here [7].
Exercises: If E = − 5 V show that 5 × E = 0. If B = 5 × A show that
5.B = 0.
There are several corollaries thatH follow from the theorems
above.
H
Corollaries from Theorem 1: i) E.dl = 0 and ii) E.dl is independent
of the integration path for given endpoints.
H
H
Corollaries from Theorem 2: i) B.dA = 0 and ii) B.dA is independent
of the integration surface for a given boundary line.
Exercise: Verify the above corollaries.
2.8
Curvilinear Coordinates
Before discussing curvilinear coordinates let us first review rectangular coordinates. Some references for this section are [2, 12, 13].
2.8.1
Plane Cartesian (Rectangular) Coordinates
For plane rectangular coordinates the î and ĵ unit vectors lie along the
x, y axes as shown on Fig. 2.7. Thus any vector A is written as A =
Ax î + Ay ĵ where (Ax , Ay ) are the x and y components. Let dlx and dly
be infinitessimal increments of length obtained when moving in the î and ĵ
directions respectively. Obviously we then have
dlx = dx
(2.46)
dly = dy
(2.47)
and
The infinitessimal displacement vector is then
dl = dlx î + dly ĵ = dxî + dy ĵ.
(2.48)
The infinitessimal area element (magnitude only) is
dA ≡ dlx dly = dxdy.
(2.49)
in plane rectangular coordinates. Thus the area of a rectangle (for which
plane rectangular
coordinates
are eminently appropriate) is given by Area
R
RL RW
of Rectangle = dA = 0 dx 0 dy = LW . That is, the area of a rectangle
equals length times width.
38
2.8.2
CHAPTER 2. VECTORS
Three dimensional Cartesian Coordinates
For rectangular coordinates the î, ĵ, k̂ unit vectors lie along the x, y, z axes
as shown in Fig. 2.8. Thus any vector is written A = Ax î + Ay ĵ + Az k̂. Let
dlx , dly , dlz be infinitessimal increments of length obtained when moving in
the î, ĵ, k̂ directions respectively. Obviously we then have
dlx = dx
(2.50)
dly = dy
(2.51)
dlz = dz.
(2.52)
The infinitessimal volume element is
dτ ≡ dlx dly dlz = dxdydz.
(2.53)
in plane rectangular coordinates. Thus for example the volume of a cube (for
which plane rectangular
coordinates
are eminently appropriate) is given by
R
R
RW RH
Volume of Cube = dτ = L
dx
dy
0
0
0 dz = LW H. That is, the volume
of a cube equals length times width times height.
The reason for going through this analysis with rectangular coordinates
is to shed light on the results for curvilinear coordiantes to which we now
turn.
2.8.3
Plane (2-dimensional) Polar Coordinates
Before discussing these coordinates let us first consider how the coordinates
of a point P (x, y) (or the components of a vector A) change to P (x0 , y 0 )
when the axis system is rotated by angle θ as shown in Fig. 2.10. Upon
examination of the figure we have
x0 = x cos θ + y sin θ
y 0 = y cos θ − x sin θ
or
or, inverted as
Ã
Ã
x0
y0
x
y
!
Ã
=
!
Ã
=
cos θ sin θ
− sin θ cos θ
cos θ
sin θ
− sin θ
cos θ
(2.54)
!Ã
!Ã
x
y
x0
y0
!
(2.55)
!
(2.56)
2.8. CURVILINEAR COORDINATES
39
Note that a counterclockwise rotation is defined to be positive. These results
will be used extensively. Let us now discuss plane polar coordinates.
This 2-dimensional coordinate system is specified by radial (r) and angular (θ) coordinates together with unit vectors (êr , êθ ) shown in Fig. 2.9. (θ
varies from 0 to 2π). Whereas for rectangular coordinates the basis vectors
are fixed in position, now with plane polar coordiantes the basis vectors are
attached to the moving point P and êr , êθ move as P moves. The relation
between plane polar and plane rectangular coordinates is obviously
x = r cos θ
y = r sin θ
(2.57)
or
r2 = x2 + y 2
θ = arctan(y/x)
(2.58)
Just as a vector A is written in rectangular coordinates as A = Ax î + Ay ĵ,
so too in polar coordinates we have
A = Ar êr + Aθ êθ
(2.59)
where (Ar , Aθ ) are the radial and angular coordinates respectively. Clearly
Ar and Aθ can be thought of as the x0 , y 0 coordinates introduced above.
Thus
Ã
! Ã
!Ã
!
Ar
cos θ sin θ
Ax
=
(2.60)
− sin θ cos θ
Aθ
Ay
and so A = Ar êr + Aθ êθ = (Ax cos θ + Ay sin θ)êr + (−Ax sin θ + Ay cos θ)êθ
= Ax î + Ay ĵ. Equating coefficients of Ax and Ay respectively, we have
î = cos θêr − sin θêθ and ĵ = sin θêr + cos θêθ or
Ã
or
Ã
î
ĵ
êr
êθ
!
Ã
=
!
Ã
=
cos θ
sin θ
− sin θ
cos θ
cos θ sin θ
− sin θ cos θ
!Ã
!Ã
êr
êθ
î
ĵ
!
(2.61)
!
.
(2.62)
That is, the relation between the unit vectors is
êr = cos θî + sin θĵ
êθ = − sin θî + cos θĵ.
(2.63)
40
CHAPTER 2. VECTORS
Let dlr and dlθ be infinitessimal increments of length obtained when
moving in the êr and êθ directions respectively. From Fig. 2.9 we have
dlr = dr
(2.64)
dlθ = rdθ
(2.65)
and
as shown in Fig. 2.11. The infinitessimal displacement vector is then
dl = dlr êr + dlθ êθ = drêr + rdθêθ
(2.66)
The circumference of a circle (for
whichR plane polar coordinates are emiR
nently appropriate) is given by dlθ = 2π
0 rdθ = 2πr. The infinitessimal
area element (magnitude only) is
dA ≡ dlr dlθ = rdrdθ
so that the area of a circle is
πR2 .
2.8.4
R
dA =
R
dlr dlθ =
(2.67)
RR
0
rdr
R 2π
0
dθ = ( 12 R2 )(2π) =
Spherical (3-dimensional) Polar Coordinates
This coordinate system is specified by (r, θ, φ) and unit vectors (êr , êθ , êφ )
shown in Fig. 2.12. In order to sweep out a sphere φ varies from 0 to 2π,
but θ now only varies from 0 to π. (Note that êθ points ’down’ whereas for
plane polar coordinates it pointed ’up’.) Again the basis vectors move as
point P moves. The relation between polar and rectangular coordinates is
obtained from Fig. 2.13 as
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
(2.68)
where r2 = x2 + y 2 + z 2 . The unit vectors are related by
êr = sin θ cos φî + sin θ sin φĵ + cos θk̂
êθ = cos θ cos φî + cos θ sin φĵ − sin θk̂
êφ = − sin φî + cos φĵ
(2.69)
2.8. CURVILINEAR COORDINATES
41
or



êr
cos θ cos φ

 
 êθ  =  sin θ cos φ
− sin φ
êφ


î
− sin θ


cos θ
  ĵ  .
0
k̂
cos θ sin φ
sin θ sin φ
cos φ
(2.70)
(do Problem 2.13) Any vector A is written as A = Ar êr + Aθ êθ + Aφ êφ .
Let dlr , dlθ , dlφ be infinitessimal increments of length obtained when moving
in the êr , êθ and êφ directions respectively. Clearly
dlr = dr
(2.71)
dlθ = rdθ
(2.72)
dlφ = r sin θdφ
(2.73)
and
but
which can be seen from Fig. 2.14. The infinitessimal displacement vector is
then
dl = dlr êr + dlθ êθ + dlφ êφ = drêr + rdθêθ + r sin θdφêφ
(2.74)
There are three infinitessimal area elements that we can form, namely dA0 =
dlr dlθ = rdrdθ (variable r) or dA00 = dlr dlφ = r2 sin θdrdφ (variable r) and
the one which has fixed r and gives the area patch on the surface of a sphere
dA = dlθ dlφ = r2 sin θdθdφ.
R
Thus the surface area of a sphere is dA =
4πR2 . The infinitessimal volume element is
R
(2.75)
dlθ dlφ = R2
Rπ
0
sin θdθ
R 2π
0
dτ = dlr dlθ dlφ = r2 sin θdrdθdφ.
so that the volume of a sphere is
= 43 πR3 .
2.8.5
R
dτ =
R
dlr dlθ dlφ =
dφ =
(2.76)
RR
0
r2 dr
Rπ
0
sin θdθ
R 2π
0
dφ
Cylindrical (3-dimensional) Polar Coordinates
These coordinates specified by (ρ, φ, z) and (êρ , êφ , k̂) are shown in Fig 2.15.
It is worthwhile to note that for a fixed slice in z, cylindrical polar coordinates
are identical to plane polar coordinates. The angle θ and radius r used in
plane polar coordinates is replaced by angle φ and radius ρ in cylindrical
42
CHAPTER 2. VECTORS
polar coordinates. In other words cylindrical polar coordinates are just plane
polar coordinates with a z axis tacked on. The relation between cylindrical
polar coordinates and rectangular coordinates is thus
x = ρ cos φ
y = ρ sin φ
z=z
(2.77)
the first two of which are analagous to equation (2.57). Unit vectors are
related by
êρ = cos φî + sin φĵ
êφ = − sin φî + cos φĵ
êz = k̂
(2.78)
again the first two of which are just equation (2.63). Any vector is
A = Aρ êρ + Aφ êφ + Az êz .
(2.79)
Note that for spherical polar coordinates the position vector of point P is
r = rêr and for plane polar coordinates r = rêr also. However in cylindrical
polar coordinates we have
r = ρêρ + zêz
(2.80)
as can be seen from Fig. 2.15. The infinitessimal elements of length are
dlρ = dρ
(2.81)
dlφ = ρdφ
(2.82)
dlz = dz
(2.83)
and
where (2.81) and (2.82) are the same as (2.64) and (2.65). The infinitessimal
displacement vector is
dl = dlρ êρ + dlφ êφ + dlz êz = dρêρ + ρdφêφ + dzêz
(2.84)
to be compared to (2.66). Exercise: derive the formula for the volume of a
cylinder.
2.9. SUMMARY
2.8.6
Div, Grad and Curl in Curvilinear Coordinates
See Griffiths.
(to be written up later)
2.9
Summary
43
44
CHAPTER 2. VECTORS
2.10
Problems
2.1 a) Vector C points out of the page and D points to the right
in the same plane. What is the direction of C × D and D × C ?
b) B points to the left and A points down the page. What is
the direction of B × A and A × B ?
2.2 a) Write down the values of the following Kronecker delta
symbols: δ11 , δ12 , δ33 , δ13 .
b) Write down the values of the following Levi-Civitá symbols:
²111 , ²121 , ²312 , ²132 .
2.3 Show that (A × B)z = Ax By − Ay Bx .
2.4 Show that the determinant formula (2.23) gives the same
results as (2.22).
2.5 Show that (2.24) is true for the following values of indices:
a) i = 1, j = 1, l = 1, m = 1,
b) i = 3, j = 1, l = 1, m = 3.
2.6 Prove the following vector identities:
a) A.(B × C) = B.(C × A) = C.(A × B)
b) A.(B × C) = −B.(A × C)
c) A × (B × C) = B(A.C) − C(A.B)
d) (A × B).(C × D) = (A.C)(B.D) − (A.D)(B.C)
e) A × [B × (C × D)] = B[A.(C × D)] − (A.B)(C × D).
2.7 Calculate the gradient of f (x, y, z) = x + yz 2 .
2.8 Calculate the divergence of C = −y î + xĵ and interpret your
result.
2.10. PROBLEMS
2.9 Calculate the curl of B = ĵ and interpret your result.
2.10
Let f = xy 2 and a = (0, 0, 0) and b = (1, 1, 0). Evaluate
Rb
a 5f.dl along two different integration paths and show that the
results are the same.
2.11 Check the fundamental theorm of gradients using the function and end points of Problem 2.10.
2.12 Check Gauss’ divergence theorem using C = xz î+y 2 ĵ +yz k̂
using the unit cube with a corner at the origin as shown in Fig.
2.6.
2.13 Prove equation (2.69) or (2.70).
45
46
CHAPTER 2. VECTORS
2.11
Answers
2.1 a) C × D points up the page and D × C points down the
page.
b) B × A points out of the page and A × B points into the page.
2.2 a) 1, 0, 1, 0.
b) 0, 0, +1, -1.
2.7 î + z 2 ĵ + 2yz k̂.
2.8 0
2.9 0
2.10 1
2.12. SOLUTIONS
2.12
47
Solutions
2.1 a) C × D points up the page and D × C points down the
page.
b) B × A points out of the page and A × B points into the page.
2.2 a) 1, 0, 1, 0.
b) 0, 0, +1, -1.
2.3 From (2.20) A × B = ²ijk Ai Bj êk . Therefore the kth component is (A × B)k = ²ijk Ai Bj . Thus (A × B)3 = ²ij3 Ai Bj =
²1j3 A1 Bj + ²2j3 A2 Bj + ²3j3 A3 Bj . Now ²3j3 = 0 for all values of j and in the first two terms the only non-zero values
will be j = 2 and j = 1 respectively. Realizing this saves
us from writing out all the terms in the sum over j. Thus
(A × B)3 = ²123 A1 B2 + ²213 A2 B1 = A1 B2 − A2 B1 meaning that
(A × B)z = Ax By − Ay Bx .
2.4
¯
¯ ê
¯ 1
¯
A × B = ¯ A1
¯
¯ B1
¯
ê2 ê3 ¯¯
¯
A2 A3 ¯
¯
B2 B3 ¯
= ê1 A2 B3 + ê2 A3 B1 + ê3 A1 B2
−ê3 A1 B2 − ê2 A1 B3 − ê1 A3 B2
= (A2 B3 − A3 B2 )ê1 + (A3 B1 − A1 B3 )ê2
+(A1 B2 − A2 B1 )ê3 .
48
CHAPTER 2. VECTORS
2.5 ²kij ²klm = δil δjm − δim δjl
a) The left hand side is
²kij ²klm = ²k11 ²k11
= ²111 ²111 + ²211 ²211 + ²311 ²311
= (0)(0) + (0)(0) + (0)(0)
= 0.
The right hand side is δ11 δ11 −δ11 δ11 = (1)(1)−(1)(1) = 1−1 = 0
b) The left hand side is
²kij ²klm = ²k31 ²k13
= ²131 ²113 + ²231 ²213 + ²331 ²313
= (0)(0) + (+1)(−1) + (0)(0)
= 0 − 1 + 0 = −1.
The right hand side is δ31 δ13 − δ33 δ11 = (0)(0) − (1)(1) = −1
2.12. SOLUTIONS
49
2.6
a)
A.(B × C) = Ak (B × C)k = Ak ²ijk Bi Cj
= Bi ²ijk Cj Ak = Bi ²jki Cj Ak = Bi (C × A)i = B.(C × A)
Also this is
= Cj ²ijk Ak Bi = Cj ²kij Ak Bi = Cj (A × B)j = C.(A × B).
b)
A.(B × C) = Ak (B × C)k = Ak ²ijk Bi Cj
= Bi ²ijk Ak Cj = Bi ²jki Ak Cj = −Bi ²kji Ak Cj
= −Bi (A × C)i = −B.(A × C).
c)
A × (B × C) = ²ijk Ai (B × C)j êk = ²ijk Ai ²jlm Bl Cm êk
= ²ijk ²jlm Ai Bl Cm êk = −²ikj ²jlm Ai Bl Cm êk
= −(δil δkm − δim δkl )Ai Bl Cm êk
= −Ai Bi Ck êk + Ai Bk Ci êk
= −C(A.B) + B(A.C).
d)
(A × B).(C × D) = (A × B)k (C × D)k
= ²ijk Ai Bj ²lmk Cl Dm
= ²ijk ²lmk Ai Bj Cl Dm
= ²kij ²klm Ai Bj Cl Dm
= (δil δjm − δim δjl )Ai Bj Cl Dm
= Ai Bj Ci Dj − Ai Bj Cj Di
= (A.C)(B.D) − (A.D)(B.C).
e)
A × [B × (C × D)] = ²ijk Ai [B × (C × D)]j êk
50
CHAPTER 2. VECTORS
= ²ijk Ai ²lmj Bl (C × D)m êk = ²ijk Ai ²lmj Bl ²stm Cs Dt êk
= ²ijk ²lmj Ai Bl ²stm Cs Dt êk = ²jki ²jlm Ai Bl ²stm Cs Dt êk
= (δil δjm − δim δjl )Ai Bl ²stm Cs Dt êk
= Ai Bk ²sti Cs Dt êk − Ai Bi ²stk Cs Dt êk
= BAi ²sti Cs Dt − (A.B)²stk Cs Dt êk
= BAi (C × D)i − (A.B)(C × D)
= B[A.(C × D)] − (A.B)(C × D).
2.13. FIGURE CAPTIONS FOR CHAPTER 2
2.13
Figure captions for chapter 2
Fig. 2.1 Sketch of the vector field A(x, y) = xî + y ĵ.
Fig. 2.2 Sketch of the vector field B = ĵ.
Fig. 2.3 Sketch of the vector field C(x, y) = −y î + xĵ.
Fig. 2.4 A multidimensional line integral with the same end
points (a, b) can be performed over different paths.
Fig. 2.5 Integration paths used in Example 2.6.1.
Fig. 2.6 Integration areas used in Example 2.6.3.
Fig. 2.7 Plane (Rectangular) Coordinates and Basis Vectors.
Fig. 2.8 3-dimensional Cartesian Coordinates and Basis Vectors.
Fig. 2.9 Plane Polar Coordinates and Basis Vectors.
Fig. 2.10 Rotation of a coordinate system.
Fig. 2.11 An increment of length dlθ moving in the êθ direction.
Fig. 2.12 Spherical Polar Coordinates and Basis Vectors.
Fig. 2.13 Relating Spherical Polar Coordinates to Cartesian Coordinates.
Fig. 2.14 An increment of length dlφ moving in the êφ direction.
Fig. 2.15 Cylindrical Polar Coordinates and Basis Vectors. (Note
that the coordinates and basis vectors in the x, y plane are identical to the plane polar coordinates and basis vectors of Fig. 2.9.)
51
52
CHAPTER 2. VECTORS
Chapter 3
MAXWELL’S EQUATIONS
This is a book about the subject of classical electrodynamics. The word
classical as used in physics does not mean ’old’ or ’pre-twentieth century’ or
’non-relativistic’ (as many students think). The word classical is very specific
and simply means that the theory is non quantum-mechanical. Actually
classical physics is still an active area of research today particularly those
branches dealing with such topics as fluid dynamics and chaos [14]. In fact
one of the major unsolved problems in classical physics is the theory of
turbulence (reference ?).
Classical electrodynamics is based entirely on Maxwell’s equations. In
fact one could define classical electrodynamics as the study of Maxwell’s
equations. Contrast this to the theory of Quantum Electrodynamics which
results when one considers the quantization of the electromagnetic field [15].
Maxwell’s equations are a set of four fundamental equations that cannot
be derived from anywhere else. They are really a guess as to how nature
behaves. The way we check whether they are correct is by comparing their
predictions to experiment. They are analagous to say Newton’s laws in
classical mechanics, or Schrodinger’s equation in quantum mechanics, or the
Einstein field equations in general relativity, all of which are not derivable
from anywhere else but are considered the starting postulates for the theory.
All we can do is to write down the postulated equation and calculate the
consequences. Of course it is always fascinating to study how these equations
were originally guessed at and what experiments or concepts lead to them
being proposed [16].
The theory of classical electrodynamics was developed in piecemeal fashion [16, 9] with the establishment of Coulomb’s law, the Biot-Savart law,
53
54
CHAPTER 3. MAXWELL’S EQUATIONS
Gauss’ law, Faraday’s law, Ampère’s law etc. However it was James Clerk
Maxwell, who in 1861, unified all the previous work, made some significant corrections of his own, and finally wrote down what are today called
Maxwell’s equations.
3.1
Maxwell’s equations in differential form
Maxwell’s equations (in vacuum ) consist of Gauss’ law for the electric field
E,
5.E = 4πkρ,
(3.1)
Gauss’ law for the magnetic field B,
5.B = 0,
(3.2)
Faraday’s law
∂B
=0
∂t
(3.3)
4πk
1 ∂E
= 2j
2
gc ∂t
gc
(3.4)
5×E+g
and Ampère’s law
5×B−
dq
) and j is the
where ρ is the charge density (charge per unit volume ρ ≡ dτ
di
current density (charge per unit area j ≡ dA Â). k and g are constants and
c is the speed of light in vacuum.
The Lorentz force law is
F = q(E + gv × B)
(3.5)
which gives the force F on a particle of charge q moving with velocity v in
an electromagnetic field.
Later we shall see that the constant k is the same one that appears in
Coulomb’s law for the electric force between two point charges
F=k
q1 q2
r̂
r2
(3.6)
and the constant g specifies the relative strength of the E and B fields. An
excellent and more complete discussion of units may be found in the book
by Jackson [10]. In terms of Jackson’s constants (k1 and k3 ) the relation
is k = k1 and g = k3 . From Coulomb’s law it can be seen that the units
3.1. MAXWELL’S EQUATIONS IN DIFFERENTIAL FORM
55
chosen for charge and length will determine the units for k. The three main
systems of units in use are called Heaviside-Lorents, CGS or Gaussian and
MKS or SI. The values of the constants in these unit systems are specified
in Table 3.1 below.
Heaviside-Lorentz
k
g
1
4π
1
c
CGS (Gaussian)
1
SI (MKS)
1
c
1
1
4π²0
Table 3.1
Inserting these constants into Maxwell’s equations (3.1) - (3.4) and the
Lorentz force law gives the equations as they appear in different unit systems.
In Heaviside-Lorentz units Maxwell’s equations are
5.E = ρ
(3.7)
5.B = 0
(3.8)
1 ∂B
=0
c ∂t
1 ∂E
1
5×B−
= j
c ∂t
c
and the Lorentz force law is
5×E+
1
F = q(E + v × B).
c
(3.9)
(3.10)
(3.11)
In CGS or Gaussian units Maxwell’s equations are
5.E = 4πρ
(3.12)
5.B = 0
(3.13)
1 ∂B
=0
c ∂t
1 ∂E
4π
5×B−
=
j
c ∂t
c
5×E+
(3.14)
(3.15)
56
CHAPTER 3. MAXWELL’S EQUATIONS
and the Lorentz force law is
1
F = q(E + v × B).
c
(3.16)
In MKS or SI units Maxwell’s equations are
ρ
²0
(3.17)
5.B = 0
(3.18)
5.E =
∂B
=0
∂t
1
1 ∂E
= 2 j
5×B− 2
c ∂t
c ²0
5×E+
However, because c2 =
is usually written
1
µ0 ²0
(3.19)
(3.20)
(see Section x.x) in SI units this last equation
5 × B − µ0 ²0
∂E
= µ0 j
∂t
(3.21)
and the Lorentz force law is
F = q(E + v × B).
(3.22)
Particle physicists [17] most often use Heaviside-Lorentz units and furthermore usually use units in which c ≡ 1, so that Maxwell’s equations are in
their simplest possible form. CGS units are used in the books by Jackson [10]
and Ohanian [9] and Marion [11], whereas MKS units are used by Griffiths
[2] and most freshman physics texts. In this book we shall use Maxwell’s
equations as presented in equations (3.1) - (3.4) so that all of our equations
will contain the constants k and n. The equations for a specific unit system
can then simply be obtained by use of Table 3.1. Thus this book will not
make a specific choice of units. The advantage of this is that comparison of
results in this book with the references will be made much easier.
3.2
Maxwell’s equations in integral form
In freshman physics course one usually does not study Maxwell’s equations
in differential form but rather one studies them in integral form. Let us now
prove that the Maxwell’s equations as presented in equations (3.1) - (3.4) do
in fact give the equations that one has already studied in freshman physics.
3.3. CHARGE CONSERVATION
57
To accomplish this we perform a volume integral over the first two equations
and an area integral over theR second two equations.
Integrating over volume
( dτ ) on Gauss’
law for E (3.1)
and using Gauss’
R
H
R
divergence theorem as in (5.E)dτ = E.dA, and q = ρdτ yields
Φ0E
≡
I
E.dA = 4πkq
(3.23)
which is Gauss’ law for the electric flux Φ0E over a closed surface area. The
magnetic equation (3.2) similarly becomes
Φ0B ≡
I
B.dA = 0
(3.24)
where Φ0B is the magnetic flux
over a closed surface area.
R
Integrating Rover area ( dA)
on Faraday’s law and using Stokes’ curl
H
theorem, as in (5 × E).dA = E.dl, yields
I
E.dl + g
∂ΦB
=0
∂t
(3.25)
H
where ΦB ≡ B.dA is the magnetic flux (not necessarily over a closed
surface
area). Finally integrating Ampère’s law over an area and using i ≡
R
j.A, yields
I
1 ∂ΦE
4πk
B.dl − 2
(3.26)
= 2i
gc ∂t
gc
H
where ΦB ≡ B.dA is the electric flux (not necessarily over a closed surface
area).
This completes our derivation of Maxwell’s equations in integral form.
Exercise: Using Table 3.1, check that the equations above are the equations that you studied in freshman physics.
3.3
Charge Conservation
Conservation of charge is implied by Maxwell’s equations. Taking the diver∂
gence of Ampère’s law gives 5.(5 × B) − gc12 ∂t
5 .E = 4πk
5 .j. However
gc2
5.(5 × B) = 0 (do Problem 3.1) and using the electric Gauss law we
∂ρ
have − 4πk
= 4πk
5 .j. We see that the constants cancel leaving
gc2 ∂t
gc2
5.j +
∂ρ
=0
∂t
(3.27)
58
CHAPTER 3. MAXWELL’S EQUATIONS
which is the continuity equation, or conservation of charge in differential
form. Because the constants all cancelled, this equation is the same in all
systems of units. This equation is a local conservation law in that it tells us
how charge is conserved locally. That is if the charge density increases in
some region locally (yielding a non-zero ∂ρ
∂t ), then this is caused by current
flowing into the local region by the amount ∂ρ
∂t = − 5 .j. If the charge
decreases in a local region (a negative − ∂ρ
),
then this is due to current
∂t
∂ρ
flowing out of that region by the amount − ∂t = 5.j. The divergence here
is positive corresponding to j spreading outwards (see Fig. 2.1).
Contrast this with the global conservation law
obtained by integrating
∂ R
over the volume
of the Hwhole global universe. ∂t ρdτ = ∂q
∂t where q is the
R
charge and 5.jdτ = j.dA according to Gauss’
divergence
theorem. We
H
are integrating over the whole universe and so dA covers the ’surface area’
of the universe at infinity. But byHthe time we reach infinity all local currents
will have died off to zero and so j.dA = 0 yielding
∂q
=0
∂t
(3.28)
which is the global conservation of charge law. It says that the total charge
of the universe is constant.
Finally, let’s go back and look at Ampère’s law. The original form of
j.
Ampère’s law didn’t have the second term. It actually read 5 × B = 4πk
gc2
From our above discussion this would have lead to 5.j = 0. Thus the original
form of Ampère’s law violated charge conservation [Guidry p.74]. Maxwell
added the term − ∂E
∂t , which is now called Maxwell’s displacement current.
1 ∂E
4πk
Writing jD ≡ 4πk
∂t , Ampère’s law is 5 × B = gc2 (j + jD ), or in integral
H
form B.dl = 4πk
(i + iD ) Maxwell’s addition of the displacement current
gc2
made Ampère’s law agree with conservation of charge.
3.4
Electromagnetic Waves
Just as conservation of charge is an immediate consequence of Maxwell’s
equations, so too is the existence of electromagnetic waves, the immediate
interpretation of light as an electromagnetic wave. This elucidation of the
true nature of light is one of the great triumphs of classical electrodynamics
as embodied in Maxwell’s equations.
3.4.
ELECTROMAGNETIC WAVES
59
First let us recall that the 1-dimensional wave equation is
∂2ψ
1 ∂2ψ
−
=0
∂x2
v 2 ∂t2
(3.29)
where ψ(x, t) represents the wave and v is the speed of the wave. Contrast
this to several other well known equations such as the heat equation [5]
∂ 2 ψ 1 ∂ψ
−
=0
∂x2
v ∂t
(3.30)
or the Schrödinger equation [17]
−
h̄2 ∂ 2 ψ
∂ψ
+ V ψ = ih̄
2
2m ∂x
∂t
(3.31)
or the Klein-Gordon equation [17]
(22 + m2 )φ = 0
(3.32)
where
1 ∂2
. − 52
v 2 ∂t2
In 3-dimensions the wave equation is
22 ≡
52 ψ −
1 ∂2ψ
=0
v 2 ∂t2
(3.33)
(3.34)
We would like to think about light travelling in the vacuum of free space
far away from sources of charge or current (which actually do produce the
electromagnetic waves in the first place). Thus we set ρ = j = 0 in Maxwell’s
equations, giving Maxwell’s equations in free space as
5.E = 0,
(3.35)
5.B = 0,
(3.36)
∂B
= 0,
∂t
1 ∂E
5×B− 2
=0
gc ∂t
5×E+g
(3.37)
(3.38)
∂
(5 × B) = 0 and
Taking the curl of Faraday’s law gives 5 × (5 × E) + g ∂t
2
substituting 5 × B from Ampère’s law gives 5 × (5 × E) + c12 ∂∂tE
2 = 0.
60
CHAPTER 3. MAXWELL’S EQUATIONS
However 5 × (5 × E) = 5(5.E) − 52 E = −52 E because of (3.35). (do
Problem 3.2). Thus we have
52 E −
1 ∂2E
=0
c2 ∂t2
(3.39)
and with the same analysis (do Problem 3.3)
52 B −
1 ∂2B
=0
c2 ∂t2
(3.40)
showing that the electric and magnetic fields correspond to waves propagating in free space at speed c. Note that the constant g cancels out, so that
these wave equations look the same in all units. It turns out that the per√
meability and permittivity of free space have the values such that 1/ µ0 ²0
equals the speed of light ! Thus the identification was immediately made
that these electric and magnetic waves are light.
Probably the physical meaning of Maxwell’s equations is unclear at this
stage. Don’t worry about this yet. The purpose of this chapter was to give a
brief survey of Maxwell’s equations and some immediate consequences. We
shall study the physical meaning and solutions of Maxwell’s equations in
much more detail in the following chapters.
3.5
Scalar and Vector Potential
Remember our theorems in potential theory (Section 2.7) ? These were i)
C = − 5 φ iff 5 × C = 0 and ii)G = 5 × F iff 5.G = 0.
From the second theorem the magnetic Gauss law (3.2) immediately
implies that B can be written as the curl of some other vector A as
B = 5 × A.
(3.41)
The vector A is called the magnetic vector potential which we shall use often
in the following chapters. Notice that (3.41) is the same in all units.
Looking at Maxwell’s equations we don’t see any curls equal to zero
and so it looks like we won’t be using the first theorem. But wait. We
have just seen that B = 5 × A, so let’s put this into Faraday’s law as
∂
5 × E + g ∂t
(5 × A) = 0 or
5 × (E + g
∂A
)=0
∂t
(3.42)
3.5. SCALAR AND VECTOR POTENTIAL
61
Now we have a vector C ≡ E + g ∂A
∂t whose curl is zero and therefore C
can be written as the gradient of some scalar function (let’s call it V ) as
C = − 5 V , from which it follows that
E=−5V −g
∂A
∂t
(3.43)
The scalar V is called the electric scalar potential, and this equations
does depend on the unit system via the appearance of the constant g.
In the next two chapters we will be studying time-independent problems,
in which all time derivatives in Maxwell’s equations are zero. The timeindependent Maxwell’s equations are
5.E = 4πkρ,
(3.44)
5.B = 0,
(3.45)
5 × E = 0,
(3.46)
and
5×B=
4πk
j,
gc2
(3.47)
implying that
E=−5V
(3.48)
only. This equation is independent of units.
We shall be using the scalar and vector potentials often in the following
chapters where their meaning will become much clearer.
62
CHAPTER 3. MAXWELL’S EQUATIONS
Chapter 4
ELECTROSTATICS
4.1
Equations for electrostatics
Electrostatics is a study the electric part of Maxwell’s equations when all
time derivatives are set equal to zero. Thus the two equations for the electric
field are Gauss’ law
5.E = 4πkρ,
(4.1)
and
5×E=0
(4.2)
from Faraday’s law yielding the scalar potential as
E≡−5V
(4.3)
(or from setting ∂A
∂t in equation (3.43)). Substituting this last result into
Gauss’ law yields Poisson’s equation for the scalar potential
52 V = −4πkρ
(4.4)
which, in source free regions (ρ = 0) gives Laplace’s equation
52 V = 0
(4.5)
In integral form, Gauss law is (equation (3.23))
Φ0E ≡
and Faraday’s law is
I
E.dA = 4πkq
(4.6)
E.dl = 0
(4.7)
I
63
64
CHAPTER 4. ELECTROSTATICS
which also follows by using equation (4.2) in Stoke’s
curl theorem. To obtain
R
equation (4.3) in integral
form we integrate over dl and use the fundamental
R
theorem of gradients Pa (5V ).dl = V (P ) − V (a) to give
Z
V (P ) − V (a) = −
P
E.dl
(4.8)
a
Notice from (4.3) that if we add a constant to the potential (V − > V +
constant) then the value of the electric field doesn’t change. (Adding this
constant is called a gauge transformation. Maxwell’s equations are invariant
under gauge transformations.) Being able to add a constant means that
I can define the zero of potential to be anywhere I like. Thus let’s define
V (a) ≡ 0 so that
Z
V (P ) = −
P
E.dl
(4.9)
θ
where P is the point where the potential is being evaluated and a ≡ θ is now
a special point where the potential is zero. Typically the point θ is either at
the origin or at infinity. Thus (4.9) is the integral version of (4.3).
Finally let us obtain the integral form of Poisson’s equation as
Z
V (P ) = k
R
ρ
dτ
R
(4.10)
R
λ
σ
dl for line charges or k R
dA for surface charges, wehere λ and σ
or k R
are the charge per unit length and charge per unit area respectively. The
distance R is shown in Fig. 4.1 and is defined as
R ≡ r − r0
(4.11)
where r is the displacement from the origin to point P and r0 is the displacement from the origin to the charge distribution. Thus r is the displacement
from the charge to point P . Notice that we didn’t yet prove equation (4.10)
like we did with the other equations. A proper solution of Poisson’s equation
involves Green functions [8] which are beyond the scope of this book. However, we shall provide a proper justification for equation (4.10) in Section
4.xx.
It is very useful to collect and summarize our results for electrostatics.
Our basic quantities are the charge density ρ, the electric field E and the
scalar potential V .
The relation between E and ρ is
4.1. EQUATIONS FOR ELECTROSTATICS
65
5.E = 4πkρ
H
5.E = 4πkρΦ0E ≡
E.dA = 4πkq.
(4.12)
The relation between E and V is
E≡−5V
RP
V (P ) = −
θ
E.dl.
(4.13)
The relation between V and ρ is
52 V = −4πkρ
V (P ) = k
R
ρ
R dτ.
(4.14)
These relations are also summarized in Fig. 2.35 of the book by Griffiths [2].
Having established our basic equationsd for electrostatics, now let’s investigate their solutions and physical meanings.
66
4.2
CHAPTER 4. ELECTROSTATICS
Electric Field
Force is a quantity that involves two bodies (force between two charged
particles, or force between Earth and Moon), whereas field is a quantity
pertaining to one object only. Coulomb’s law for the electric force betweeen
two point charges is F = k q1r2q2 r̂ where r is the distance between the charges.
The definition of electric field gets rid of one of these charges by dividing it
out as
F
E≡
(4.15)
q
so that the electric field surrounding a single point charge is E = k rq2 r̂.
In order to calculate electric fieldss
it is much more conveneient to use
H
Gauss’ law in integral form Φ0EH ≡ E.dA = 4πkq rather than the differential version. The integral dA forms a closed surface, which is often called a Gaussian surface which fully encloses the charge q. The key
trick in using Gauss’ law is to always choose a Gaussian surface so that
E and dA are parallel or perpendicular to each other, because then either
E.dA = EdA cos 0o = EdA or E.dA = EdA cos 90o = 0 and one won’t have
to worry about vectors. The way to ensure that E and dA are parallel or
perpendicular is to choose the Gaussian surface to have the same symmetry
as the charge distribution. We illustrate this in the following examples.
————————————————————————————————–
Example 4.2.1 Prove Coulomb’s law.
Solution Coulomb’s law gives the force betweeen two point charges.
Thus we first need the electric field due to one point charge. We
know that the field surrounding a point charge is as shown in
Fig. 4.2. The field points radially outward. The only surface for
which dA points radially outward also is the sphere (also shown
in the figure) which has the same symmetry as the point charge.
If we had drawn a cube as our Gaussian surface (which would
stilll give the correct answer, but only with a much more difficult
calculation) then E and dA would not be parallel everywhere.
(Exercise: draw such a picture to convince yourself.)
Also note that the Gaussian surface has been drawn to intersect the observation point P . We always do this, so that the
distance to the observation point
is the same as the size of the
H
0
Gaussian surface. Now ΦE ≡ E.dA is easy. For a sphere dA =
r2 sin θdθdφ from equation (2.75). Note that on this Gaussian
4.2. ELECTRIC FIELD
67
sphere the magnitude of the electric field is constant
and therefore
R 2
0
may be taken outside the integral to give ΦE = E r sin θdθdφ =
E4πr2 (as shown at the end of Section 2.8.4). Thus
q
R2
(4.16)
q
R̂
R2
(4.17)
E=k
or, form our picture
E=k
and using the definition of electric field E ≡
F=k
F
q
gives
q1 q2
R̂
R2
(4.18)
Thus we see how Gauss’ law yields Coulomb’s law.
Example 4.2.2 Find the electric field due to a charged plate of
infinite area.
Solution The electric field lines due to a positively charge infinite
plate are shown in Fig 4.3 together with a Gaussian surface of the
same symmetry, which is a cylinder, but a square box would have
done equally well. The angle between dA3 (the area vector of the
rounded edge) and E is 90o and therefore E.dA3 = 0. Both dA2
and dA1 are parallel to E so that E.dA2 = EdA2 and E.dA1 =
EdA1 . Furthermore, on the areas A1 and A2 the electric field is
constant (independent Hof R) so that
E may be
taken outside
the
R
R
R
0 ≡
integral
to
give
Φ
E.dA
=
E.dA
+
E.dA
+
E.dA
1
2
3
R
RE
= E dA1 + E dA2 + 0 = 2EA where A is the area of the
circular ends. Thus 2EA = 4πkq. The same area A intersects
the charged plate so that we can define the surface charge density
σ ≡ Aq as the charge per unit area giving
E = 2πkσ
(4.19)
E = 2πkσ R̂
(4.20)
or
Note that here the electric field is independent of the distance
from the plate.
————————————————————————————————–
68
CHAPTER 4. ELECTROSTATICS
Exercise: A) Explain why the electric field is independent of the distance
from the plate. B)Infinite plates don’t exist. Why is our problem still of
practical importance ? C) A capacitor consists of a positive plate and a
negative plate separated by a distance d. What is the electric field inside
and outside the capacitor ?
(do Problems 4.1, 4.2, 4.3)
4.3
Electric Scalar potential
In classical mechanics one always has the option of solving problems with
either force methods or energy methods. The force methods involve the use
of Newton’s law F = ma, while the energy methods use conservation of
energy. (The sophisticated version of energy methods is expressed in the
re-formulations of mechanics via Lagrangian and Hamiltonian dynamics).
In electrodynamics ’force’ methods involve calculation of electric field
(trivially related to force via E ≡ Fq ). Alternatively the ’energy methods’
are based upon calculation of the electric potential V . If we have the electric
potential it is then easy to get the field just from E = − 5 V .
A word on naming things. The force betweeen two charged particle
is F = k q1r2q2 r̂ and the potential energy for a conservative force is always
F = − 5 U giving U = k q1rq2 as the potential energy between two point
charges. We defined the electric field intrinsic to a single chagre as E ≡ Fq
and similarly we can define (actually it follows) the potential V ≡ Uq which is
sort of like the ’energy’ intrnsic to a single charge. (Note that in MKS units,
the potential has units of volts, so the V symbol is a good one. Actually a
better name for V is voltage rather than potential, because then we don’t
get mixed up with potential energy.)
In mechanics it is very often useful to have both the concept of force and
the concept of energy. Similarly in electrodynamics it turns out to be very
useful to have both the concept of field and the concept of potential. We
shall illustrate the use of p[otential in the following example.
————————————————————————————————–
Example 4.3.1 Calculate the electric potential inside and outside a uniformly charged spherical shell of radius a. A) Use equation (4.13). B) Use equation (4.14). (See examples 6 and 7 in
chapter 2 of the book by Griffiths [2].)
Solution A) In problem 4.1 the electric field was calculated as
4.3. ELECTRIC SCALAR POTENTIAL
69
E(r < a) = 0 inside the sphere and E(r > a) = k rq2 r̂ ≡ k rq2 êr
outside. Let us use equation (4.13) in the form
Z
V (P ) = −
P
∞
E.dl
(4.21)
where we have choen the reference point of zero potential at
infinity, i.e. V (∞) = 0.
In spherical polar coordinates dl = dlr êr + dlθ êθ + dlφ êφ , so that
for points outside the sphere E.dl = k rq2 êr .(drêr + dlθ êθ + dlφ êφ )
= k rq2 dr giving
Z R
1
kq
dr =
−kq
(4.22)
2
R
∞r
This is also the potential of a point charge.
Notice that we have integrated all the way out from ∞ up to the
point P at R, but we didn’t crosss the surface of the sphere in
our journey because the point P at R is outside the sphere.
Continuing our journey across the spherical boundary and inside
the sphere gives us the potential inside. But now E has two
different values depending
upon whether
we are inside or outside
R
R
so that V (r < a) = −[ a∞ E.dl + ra E.dl] but the second integral
is zero because E(r < a) = 0. The first integral is just the one
we did above except the upper limit of integration is a rather
than R. Thus V (r < a) = kq
a showing that the potential is not
zero inside the sphere but rather is a constant. Using E = − 5 V
does give zero electric field inside the sphere.
B) This part of the problem shows how to calculate the potential
if we haven’t already got the electric field E. The solution
to
R
Poisson’s equation is given in equation (4.14) as V (P ) = k Rρ dτ .
It is important to note that R is the vector from a point within
the charge distribution to the observation point P , whereas dτ
is an integral over the charge density ρ only. (See Fig. 4.1.) An
easy understanding of equation (4.14) is to consider it simply as
a generalization of equation (4.22). For a surface charge density
σ (charge per unit area), ρdτ is simply replaced with σdA.
To make things easy for us let’s put the point P on the z axis and
the origin at the center of the sphere as shown in Fig. 4.4. Using
the law of cosines we have R2 = z 2 + a2 − 2za cos θ. In spherical
70
CHAPTER 4. ELECTROSTATICS
polar coordinates the appropriate area element is dA = dlθ dlφ
= a2 sin θdθdφ to give
V (z) = kσ
R
2
√ a sin θdθdφ
z 2 +a2 −2za cos θ
= 2πa2 kσ
Rπ
0
The integral is easy to do, giving
V (z) =
p
2πakσ
z [
p
(z + a)2 − (z − a)2 ] =
√
sin θdθ
.
z 2 +a2 −2za cos θ
p
2πakσ
z [
p
(z + a)2 − (a − z)2 ].
For points outside the sphere (z > a) we use the first expression
(taking only the positive roots) to give
2
4πa kσ
V (z > a) = 2πakσ
= kq
z [(z +a)−(z −a)] =
z
z (where we have
q
2πakσ
used σ = 4πa2 and inside V (z < a) = z [(z + a) − (a − z)] =
4πakσ = kq
a .
Of course the orientation of our axes is arbitrary so that we may
kq
kq
replace V (z > a) = kq
z by V (R > a) = R and V (z < a) = a
by V (R < a) = kq
a in agreement with our results above. (do
Problems 4.4 and 4.5).
————————————————————————————————–
4.4
Potential Energy
Work is defined as
W ≡
Z
F.dl
(4.23)
and substituting the right hand
side ofR F = ma the
integral (let’s
just
R
R dv dx
R dv
dv
do 1 dimension) becomes m adx = m dt dx = m dx dt dx = m v dx dx
R
= m fi vdv = ∆T where the kinetic energy is defined as T ≡ 12 mv 2 and
∆T ≡ Tf − Ti . Thus work is always equal to the change in kinetic energy.
Now let’s break the work up into a conservative (C) and a non-conservative
(N C) piece as
W ≡ WC + WN C =
Z
F.dl = ∆T
Z
=
FC .dl + WN C
(4.24)
and let’s define the conservative piece of the force (F = FC + FN C ) as
FC ≡ − 5 U
(4.25)
4.4. POTENTIAL ENERGY
71
where U is the potential energy (as opposed to the potential or voltage).
This relation between conservative force and potential energy is analagous
to the relation between electric field and potential (voltage), E = − 5 V .
From potential theory it immediately follows that
5 × FC = 0,
I
FC .dl = 0.
=>
(4.26)
(In electromagnetic theory we started with 5 × E = 0 and then got E =
− 5 V . Above we are starting with a definition FC ≡ − 5 U and then get
5 × FC = 0.) The conservative part of the work becomes
Z
WC =
Z
FC .dl ≡ −
f
i
5 U.dl = −(Uf − Ui ).
(4.27)
Thus
WC = −∆U
(4.28)
from the fundamental theorem of gradients. Substituting back into (4.24)
we have −∆U + WN C = ∆T or
∆U + ∆T = WN C
(4.29)
which is the famous work-energy theorem. WN C stands for non-conservative
work such as heat, sound energy etc. If no heat or sound is involved then
the mechanical energy (T + V ) is conserved via ∆U + ∆T = 0.
Equation (4.29) is also the First Law of Thermodynamics. To see this
write WN C = ∆Q for the change in heat and W = ∆T (remember it’s always
this). Then ∆U + ∆W = ∆Q which is more often written as
dU + dW = dQ
(4.30)
We already have a formula for the kinetic energy as T = 12 mv 2 which
is always true. What about a formula for the potential energy U ? Well
our formula for U will depend on the force. Recall that the formula for T
came from integrating the right hand side of F = ma and U comes from
the left hand side. That is why T is always the same, 12 mv 2 , but U changes
depending on what F is.
72
CHAPTER 4. ELECTROSTATICS
In what we study below we are interested in how much work I must do
against a force. Thus
F.dl = F dl cos θ = −F dl
(4.31)
because θ = 180o if I am pulling in the direction dl but the force is pulling in
the opposite direction. Having decided upon the signs, then F in the above
equation is now a scalar . For example, for a spring we often write F = −kx,
but since I have already dealt with signs I use F = kx in equation (4.31).
For all the cases below the conservative work is
WC = −∆U =
Z
giving
f
i
Z
F.dl = −
Z
f
F dl
(4.32)
i
f
F dl
∆U =
(4.33)
i
Gravity near the surface of Earth (F = mg). Here we have
Uf − Ui = mg
Z
f
dl
i
= mglf − mgli
(4.34)
and leaving off the i and f symbols yields
U = mgl
(4.35)
in general, which is just the formula for the potential energy that we learnt in
freshman physics (potential energy = mgh). Note that the zero of potential
energy is at l = 0, i.e. U (l = 0) = 0.
Simple Harmonic Oscillator. (F = kx).
Uf − Ui = k
Z
f
xdx
i
1
1
= kxf 2 − kxi 2
2
2
(4.36)
yielding
1
U = kx2
2
where the zero of potential energy is U (x = 0) = 0.
(4.37)
4.4. POTENTIAL ENERGY
73
Universal Gravitation. (F = G m1r2m2 ).
Z
f
dr
2
i r
m1 m2
m1 m2
= −G
+G
rf
ri
Uf − Ui = Gm1 m2
(4.38)
yielding
m1 m2
(4.39)
r
where now the zero point of potential energy is at infinity, namely U (r =
∞) = 0.
Coulomb’s Law. Here we have a problem. The sign of the force is different
depending on whether we have like or unlike charges, giving a repulsive or
attractive force respectively. The easiest way not to get confused is to copy
gravity which is always attractive. Thus the formulas for unlike charges will
look just like gravity and for like charges we just change whatever sign shows
up for gravity. Thus ’copying’ equation (4.39)
q 1 q2
U = −k
(4.40)
r
U = −G
for unlike charges (attractive force). Therefore we must have
q 1 q2
(4.41)
U = +k
r
for like charges.
All Forces. It’s a bit of a nuisance having to always go throught the effort
of the above calculations for potential energy. There is a good memory device
for getting the answer instantly. The memory device is
F = |F| = Cxn
=>
U =C
xn+1
n+1
(4.42)
which makes ’sense’ from F = −5U . In using this memory device |F| means
that we ignore any sign for the force. x is the corresponding distance.
Gravity near Earth. (F = |F| = mg). Thus n = 0 because a distance
1
does not appear in the force. Here C = mg giving U = mg l1 = mgl in
agreement with (4.34).
Simple Harmonic Oscillator. (F = |F| = kx). We have n = 1 and C = k
1+1
giving U = k x 2 = 12 kx2 in agreement with (4.36).
Universal Gravitation. (F = |F| = G m1r2m2 ). We have n = −2 and
−2+1
C = Gm1 m2 giving U = Gm1 m2 r−2+1 = −G m1rm2 in agreement with (4.38).
74
4.4.1
CHAPTER 4. ELECTROSTATICS
Arbitrariness of zero point of potential energy
The zero point of potential energy and potential (point at which they are
equal to zero) is supposed to be completely arbitrary [6] Pg. 370, yet from
the above analysis it looks like the zero point was automatically determined.
(We are working with F = − 5 U but identical results hold for E = − 5 V .
Remember E ≡ Fq and V ≡ Uq .)
We have
F=−5U
=> −∆U =
Z
F.dl
Z
Ui − Uf =
f
F.dl
(4.43)
i
Let’s choose Ui ≡ 0 at the special point i0 giving
Z
f
Uf = −
F.dl
(4.44)
i0
with U (i0 ) ≡ 0, or
Z
U (P ) = −
P
F.dl
(4.45)
i0
analagous to equations (4.13) and (4.21). With F.dl = F dl cos θ = −F dl as
we chose above we would have
Z
P
U (P ) =
F dl
(4.46)
i0
in perfect agreement with our results before. In fact this formula shows why
our memory device works. If the point i0 = ∞ then
Z
U (P ) = −
P
∞
F.dl
(4.47)
with U (∞) ≡ 0 the same as (4.21).
4.4.2
Work done in assembling a system of charges
The work done in assembling a system of charges is identical to the energy
stored in a system of charges. The potential energy stored in a system of
4.4. POTENTIAL ENERGY
75
two like charges, from equation (4.41) is U = +k q1rq2 . The work required to
assemble these two like charges is just W = U = k q1rq2 . But V ≡ Uq and so
W = qV
(4.48)
is the work required to put a single charge into potential V . The work
required to assembel a system of charges is
1X
1
W =
q i Vi =
2 i
2
Z
ρV dτ
(4.49)
where the 12 factor comes from double counting
[2], Pg. 93-94. We re-write
1 R
this using 5.E = 4πkρ to give W = 12 4πk
(5.E)V dτ . Now
5.(EV ) = (5.E)V + E.(5V ) = (5.E)V − E 2
(do Problem 4.6 ) so that
W =
1 R
8πk [
R
5.(EV )dτ + E 2 dτ ] =
1 R
8πk [
R
V E.dA + E 2 dτ ].
However the surface integral is zero [2], Pg. 95 giving
W =
1
8πk
Z
E 2 dτ
(4.50)
allspace
showing that the energy density E of the electric field varies as
E ∝ E2.
————————————————————————————————–
Example 4.3.2 Calculate the energy stored in the charged spherical shell discussed in Example 4.3.1 using both (4.49) and (4.50).
(See example 8 in chapter 2 of the book by Griffiths [2].)
R
R
Solution A) W = 12 ρV dτ = 12 σV dA for a surface charge.
The energy we are calculating is actually the work necessary to
bring a system of charges together to make the spherical shell
in the first place. Thus we want the potential V at the surface
of the shell. This is V = kq
a which is a constant. Therefore
q2
1 kq R
W = 2 a σdA = k 2a .
R
1
2
B) W = 8πk
allspace E dτ where the integration extends over all
space. Inside the sphere E = 0 and outside E = k rq2 êr giving
R
R
r sin θdrdθdφ
q
dr
= 2kq 2 ∞
E 2 = k 2 qr4 . Thus W = kq
a r 2 = k 2a in
8π
r4
agreement with our result above. (do Problem 4.7).
2
2
2
2
————————————————————————————————–
76
4.5
CHAPTER 4. ELECTROSTATICS
Multipole Expansion
Griffiths Chpt. 3 (leave out this time).
Chapter 5
Magnetostatics
We shall develop the theory in this chapter in exact analogy to our study of
electrostatics.
5.1
Equation for Magnetostatics
Magnetostatics is a study of the magnetic part of Maxwell’s equation when
all time derivatives are not equal to zero. Thus two equations for the magnetic field are Gauss’ Law
∇·B=0
(5.1)
and Ampèré’s Law
∇ × B = 4πkm j
(5.2)
where
km ≡
k
.
gc2
77
(5.3)
78
5.1.1
CHAPTER 5. MAGNETOSTATICS
Equations from Ampèré’s Law
The solution to Ampèré’s Law is just the Biot-Savart Law
B = km
R
j×R̂
dτ 0
R2
(5.4)
which is checked by taking ∇ × B and showing that the right hand side of
(5.2) is obtained. (do problem 5.1)
Ampèré’s Law in integral form is
H
B · dl = 4πkm i
(5.5)
The Biot-Savart law can always be saved to obtain the magnetic field,
but the calculations may often be difficult. If a high degree of symmetry
exists it is much easier to calculate the magnetic field using Ampèré’s law
(in integral form).
5.1.2
Equations from Gauss’ Law
Gauss’ law in integral form is
I
B·da = 0
(5.6)
but this isn’t very useful for us. Much more useful is the vector potential
B=∇×A
(5.7)
which is an immediate consequence of ∇ · B. Substituting into Ampèré’s law
yields Poisson’s equation
∇2 A = −4πkm j
5.2. MAGNETIC FIELD FROM THE BIOT-SAVART LAW
79
(5.8)
where we have assumed Coulomb gauge [?] (pg.227) in which
∇·A=0
(5.9)
For j = 0, poisson’s equation becomes Laplace’s equation ∇2 A = 0. The
solution to B = ∇ × A is obviously
A=
1
4π
R
B×R̂
dτ 0
R2
(5.10)
The reason this is obvious from comes looking at the solution (5.4) to
Ampèré’s law.
The solution to Poisson’s equation is obviously
Z
A = km
jdτ 0
R
(5.11)
The reason this is obvious comes from looking at the solution (??) to Poisson’s equation in electrostatics.
We have now extablished our basic equation for magnetostatics. now
let’s investigate their solutions and physical meaning.
5.2
Magnetic Field from the Biot-Savart Law
We shall illustrate the use of the Biot-Savart law by calculating the magnetic
field due to several current distrubutions.
R
ρdτ
Before
proceeding
however
recall
that
in
electrostatics
we
need
q
=
R
R
or σda or λd` for volume, surface and line charges respectively where ρ
is the charge per unit volume, σ is charge per
unit area and λ is charge per
R
unit length. In the Biot-Savart law we have jdτ where j is the current per
unit area, which can be written as
j = ρv
(5.12)
where ρ is charge per volume and v is the speed of the current. The
units of
R
Coulomb 1
j = ρv are Coulomb
=
which
is
current
per
area.
Thus
jdτ
will
3
2
sec
m
m
80
CHAPTER 5. MAGNETOSTATICS
have units of Amp·m which is current times length or charge times speed,
i.e.
Z
qv ∼ jdτ
(5.13)
R
R
R
Consequently for surfaces and lines we have qv ∼ jdτ ∼ kda ∼ id`
where j=current/area, k=current/length and i=current only.
————————————————————————————————–
Example 5.2.1 Calculate the magnetic field due to a steady
current in a long, thin wire.
solution A thin wire will carry a current i so that the Biot-Savart
law is
Z
i × R̂
d`
R2
Z
d` × R̂
= km i
R2
B = km
From Fig. 5.1 we see that d` × R̂ points out of the page and has
magnitude | d` × R̂R|= dl sin φ. For an infinitely long wire we will
∞
want to integrate −∞
d` but this is more clearly accomplished
by
R π/2
−π/2 dθ.
Thus let’s use θ as the angle variable instead of φ.
Thus | d` × R̂ |= d` sin φ = d` sin(π/2 − θ) = Rd` cos θ. Thus
θ
the magnitude of the magnetic field is B = km i d`Rcos
, but `
2
and θ are not independent and so we express one in terms of
the other so that d` cos θ = cosa θ dθ. However R = cosa θ so that
i R θ2
d` cos θ
= a1 cos θdθ yielding B = km
θ1 cos θdθ, giving our final
a
R2
result
km i
(5.14)
B=
(sin θ2 − sin θ1 )
a
for a wire of finite length. If the wire is infinitely long then
θ2 = π/2 and θ1 = −π/2 so that
B=
2km i
.
a
(5.15)
To obtain the direction of B, the right hand rule readily yields
that B consists of circular loops around the wire, as shown in
Fig.5.2.
————————————————————————————————–
5.3. MAGNETIC FIELD FROM AMPÈRÉ’S LAW
5.3
81
Magnetic Field from Ampèré’s Law
Ampèré’s law works similarly to Gauss’ law for electrostatics. Instead of
drawing a Gaussian surface we draw an Ampèrian loop with the same symmetry as the current distribution.
————————————————————————————————–
Example 5.3.1 Work out the previous example but using Ampèré’s
law for an infinitely long wire.
solution An Ampèrian loop drawn so that the angle between
B and d` is 0 is Hobviously a loop just like the magnetic loops
of Fig.5.2. Thus B · d` = B2πa = 4πkm i giving B = 2kam i in
agreement with (5.15). We see that Ampèré’s law gives us the
result much more quickly.
————————————————————————————————–
5.4
Magnetic Field from Vector Potential
We can also obtain the magnetic field if we have already calculated the vector
potential from B = ∇ × A, where A is calculated first using (5.11).
Example 5.4.1 A) Calculate the vector potential for the example discussed in Example 5.2.1. B) Calculate the magnetic field
from B = ∇ × A and show that it agrees with the previous result
in Example 5.2.1. (see probelm 5.25 of Griffiths)
solution See solutions in Griffiths
5.5
Units
Recall how our previous constants k and g came about. k was the arbitrary
originally obtained in Coulomb’s law F = q1r2q2 , appearing equivalently in
R
0
E = k R̂ ρdτ
or equivalently in Gauss’ law ∇ · E = 4πkρ. The constant g
R2
appeared in relating electric to magnetic phenomena via ∇ × E + g ∂B
∂t = 0
and it serves to relate E to B.
Just as Coulomb’s law is fundamental to electrostatics, so too is the BiotSavart law to magnetostatics. One might wonder therefore why a seperate
fundamental constant did not appear in the Biot-Savart law. Actually it did!
82
CHAPTER 5. MAGNETOSTATICS
R
R̂
dτ 0 .
Equation (5.4) has the constant km appearing in it as B = km j×
R2
Thus just as k is the constant fundamental to Coulomb’s law, so too km
is the constant fundamental to the Biot-Savart law. However electric and
magnetic phenomena are related and the constant g relates them. Obviously
then we have three constants but we would expect only two of them to be
truly independent. This is exemplified by our relation km ≡ gck2 in equation
(5.3). Thus we wrote Maxwell’s equations with k and g, but we could instead
have used km and k or alternatively km and g. Thus we can expand Table
3.1 into Table 5.1.
Heaviside-Lorentz
Gaussian (CGS)
SI (MKS)
k
1
4π
1
1
4π²o
g
1
c
1
c
1
1
4πc
1
c
km ≡
k
gc2
1
4π²o c2
=
µo
4π
Table 5.1.
Note that in SI units km =
vacuum.
µo
4π ,
where µo is the magnetic permeability of
Chapter 6
ELECTRO- AND
MAGNETOSTATICS IN
MATTER
6.1
Units
In this chapter we shall be introducing action fields called displacement field
D, polarization field, P, magnetization M and another field H which we
shall simply call the H field. They are related to electric and magnetic fields
by the following definitions [10]
D ≡ ²o E + λP
and
H=
1
B − λ0 M,
µo
(6.1)
(6.2)
where ²o , µo , λ and λ0 are proportionality constants [10]. D and P are
always shown to have the same units, as are B and M. The only difference
between them are the numerical factors λ and λ0 . In rationalized units (such
as Heaviside-Lorentz and SI) λ ≡ λ0 ≡ 1, whereas in unrationalized units
(such as Gaussian) λ ≡ lambda0 ≡ 4π.
However D and E need not have ths same units, nor need B and H.
(However in Heaviside-Lorentz and Gaussian system they do have the same
units.) The values of all the constants are listed in Table 6.1.
In SI units we usually just leave the symbols ²o and µo instead of writing
107
−7 [10].
their values which are ²o = 4π(c)
2 and µo = 4π × 10
83
84
CHAPTER 6. ELECTRO- AND MAGNETOSTATICS IN MATTER
Heaviside-Lorentz
Gaussian (CGS)
SI (MKS)
k
1
4π
1
1
4π²o
g
1
c
1
c
1
1
4πc
1
c
λ
1
4π
1
λ0
1
4π
1
²o
1
1
²o
µo
1
1
µo
km ≡
k
gc2
1
4π²o c2
≡
µo
4π
Table 6.1: Table 6.1
For purposes of having our equations independent of units, we shall leave
²o , µo , λ and λ0 in all equations. Note that ²o is the permittivity of vacuum
and µo is the permeability of vacuum.
As we shall discuss below, the relations for a linear medium are always
(for any unit system) [10]
D = ²E
(6.3)
B = µH
(6.4)
and
and the constants ²o , µo are the vacuum values of ², µ and according to
Jackson [10] the ratio of ²/²o is often called the relation permeability or just
the permeability.
This whole discussion in this section simply serves to introduce units.
We shall now discuss the concepts in much more detail.
6.2. MAXWELL’S EQUATIONS IN MATTER
6.2
6.2.1
85
Maxwell’s Equations in Matter
Electrostatics
Suppose a dielectric material is placed in an etermal electric field as shown
in Figure 6.1. Because the atoms and molecules within the solid contain
positive and negative charges the will become slightly polarized due to the
external field. This will lead to an overall polarization of the material with
polarization vector P which will be in a direction opposite to the external
vector E as shown in Figure 6.1. Thus the net field D in the material will
be different to E and P. This D is called the displacement field.
Just as the strength of an electric field is due to a charge density via
∇·E = 4πkρ, so too will the fields P and D be due to affective charge
densities. The polarization field will be due to the charges trapped (or
bound) within the individual microscopic dipoles. If there still remains a
residual D in the material then this will be due to non-bound or fee charges.
Thus the total charge density ρ is written in terms of free charge density
ρf and bound charge density ρb as (true in all unit systems)
ρ ≡ ρf + ρb .
(6.5)
From our discussion above there are three fields in the dielectric. We
have the external field P, so that the resultant field in the medium is D
which will be a linear combination of E and P. Thus define (true for all
units.)
D = ²o E + λP
(6.6)
where ²o and λ are, as yet, undetermined constants, both of where values
we are free to choose. The choices are listed in Table 6.1.
We would now like to find a Gauss’ law for D. Thus let’s calculate
∇ · D = ²o ∇ · E + λ · P = 4πk²o ρ + λ∇ · P
However in all unit systems (do Problem 6.1)
4πk²o = λ
(6.7)
so that
∇ · D = λ(ρ + ∇ · P)
= λρf + λ(ρb + ∇ · P).
(6.8)
(6.9)
86
CHAPTER 6. ELECTRO- AND MAGNETOSTATICS IN MATTER
Now we have ∇ · D related to ρf and ∇ · P is related to ρb . Evidently then
require (all units)
∇ · P = −ρb
(6.10)
where the minus sign can be understood because P is in a direction opposite
to E (see Figure 6.1). Thus we finally obtain Gauss’ law as (all units)
1
²o ∇
· D = 4πkρf .
(6.11)
It probably makes sense that if we increase the external electric field
then the polarization field whould increase as will. A linear medium is one
in which the polarization is directly proportional to the electtic field and
not, say, the square of the electric field. The constant of proportionality is
the susceptibility. Thus for a linear medium (for all units)
P ≡ ²o χe E
(6.12)
where χe is called the electrical susceptibility. Thus in HL and CGS units
(²o = 1) we would have P = χe E and in MKS units we have P = ²o χe E.
Substituting (6.12) into (6.11) yields (for all units)
D = ²o (1 = λχe )E ≡ ²E
(6.13)
which serves to define the permittivity ². The relative permitivitty or dielectric constant is defined as (all units)
κ≡
²
= 1 + λχe .
²o
(6.14)
Thus is HL units we have D = (1 + χe )E and κ = ² = 1 + χ. In CGS
units D = (1 = 4πχe )E and κ = ² = 1+4πχe . In MKS units D = ²(1+χe )E
and κ = ²²o = 1 + χe . Note that for vacuum we must have χe = 0.
6.2.2
Magnetostatics
We proceed in analogy with electrostatics. If an external magnetic field is
applied to a medium then a magnetization field M will be set up in the
medium analogous to the polarization field P. The resultant magnetic field
H will be a linear combination of M and B. Thus define (true for all units)
6.2. MAXWELL’S EQUATIONS IN MATTER
H≡
1
µo B
87
− λ0 M
(6.15)
where µo and λ0 are, as yet, undetermined constants, both of where values
we are free to choose. The choice are listed in Table 6.1.
We would now like to find an Ampère law for H. Thus we should calculate
∂D
∇ × H − ∂D
∂t or ∇ × H − α ∂t where α is a constant, or what? Let’s use the
original form of Ampèrés law ∇ × B − gc12 ∂E
∂t as our guide. This suggests
taking µo ∇ × H −
1 1 ∂P
.
gc2 ²o ∂t
µo ∇ × H −
Evaluating this we have
1 1 ∂D
gc2 ²o ∂t
= ∇ × B − λ0 µo ∇ × M
1 ∂E
λ ∂P
− 2
2
gc ∂t
gc ²o ∂t
4πk ∂P
4πk
j − λ0 µo ∇ × M − 2
2
gc
gc ∂t
−
=
where we have used equations (6.7) in the form 4πk = ²λo . The term ∇ × M
will correspond to bound current jb but the term ∂P
∂t will correspond to a
new [2] ”polarization” current jp . Thus we define
j ≡ j f + jb + jp
(6.16)
to give
µo ∇ × H −
1 1 ∂D
gc2 ²o ∂t
=
4πk
4πk
jf + ( 2 jb − λ0 µo ∇ × M)
2
gc
gc
4πk
∂P
+ 2 (jp −
)
gc
∂t
Now we have ∇ × H related to jf and ∇ × M related to jb and ∂P
∂t related
to jp . Thus the term in parenthesis must be zero which requires (all units)
∂P
∂t
= jp
(6.17)
and (all units)
∇×M=
1 4πk
λ0 µo gc2 jb
88
CHAPTER 6. ELECTRO- AND MAGNETOSTATICS IN MATTER
(6.18)
finally giving Ampèrés law for a medium as (all units)
µo ∇ × H −
1 1 ∂D
gc2 ²o ∂t
=
4πk
j
gc2 f
(6.19)
Finally we consider the magnetization of a linear medium. In electrostatics a linear medium was defined as one in which P ≡ ²o χe E from which it
followed that D = ²E and thus also D ∝ P and therefore any of these three
relations would serve equally well as the starting point for the definition of
a linear medium. Clearly the same should be true for magnetostatics, i.e.
H ∝ B, B ∝ M and M ∝ H. We start with (all units)
M ≡ χm H
(6.20)
where χm is called the magnetic susceptibility. Substituting this into (6.15)
yields (for all units)
B = µo (1 + λ0 χm )H ≡ µH
(6.21)
which serves to define the probability µ. The relative permiability is defined
as (all units)
µ
≡ 1 + λ0 χm
µo
6.2.3
Summary of Maxwell’s Equations
Maxwell’s equations as written in equations (31-1)-(31-4) are true always.
They are true in material medium and free space. The only trick to remember
is that in material medium, the charge density and current appearing on the
right hand side actually are ρ = ρf + ρb and j = jf + jb + jp .
We now have another set of Maxwell equations (true in all unit systems)
1
∇Ḋ = 4πkρf
²o
(6.22)
∇·B=0
(6.23)
∂P
∇×E+g
=0
(6.24)
∂t
1 1 ∂D
4πk
(6.25)
µo ∇ × H − 2
= 2 jf
gc ²o ∂t
gc
which again are true always. They are true in material media and free space.
(However they are really only useful in material media.)
6.3. FURTHER DIMENNSIONAL OF ELECTROSTATICS
6.3
89
Further Dimennsional of Electrostatics
6.3.1
Dipoles in Electric Field
As previously discussed, when materials are placed in external electric fields
they get polarized. The net polarization field P is due to the way in which
all the microscopic dipoles (such as polarized atoms or molecules) respond
to the external field.
Thus it is of interest to us to find out what happens to a dipole in an
external electri field.
The torque on a dipole in an external electric field (Pg. 162 of Griffiths
[2]) is
M=P×E
(6.26)
where N os the torque, E is the external electric field and p is the dipole
moment.
If the electric field is non-uniform there will be a net force on the dipole,
given by (Pg. 162 of Griffiths [2] )
F = (p · ∇)E
(6.27)
Finally we wish to find the energy of a dipole in an external field E. First
recall that
∇(p · E = p × (∇ × E) + E × (∇ × p) + (p · ∇)E + (E · ∇)p.
Thus ∇ × p = 0 and (E · ∇)p = 0. Also ∇ × E = 0 to give ∇(p · E) =
(p · ∇)E. Using F = −∇U where U is the potential energy, we have
U = −p · E
(6.28)
for the energy of dipole in an external field.
90
CHAPTER 6. ELECTRO- AND MAGNETOSTATICS IN MATTER
6.3.2
Energy Stored in a Dielectric
We previously found that the energy stored in an electric field (equivalently
the work required to assemble a systerm of charges is
1
W =
8πk
Z
E 2 dτ.
allspace
However for dielectric media this formula get charged to
W =
1
8πk²o
R
allspace D
· Edτ
(6.29)
R
D · Edτ and for CGS units we have
(For MKS
units we thus have W =
1 R
W = 8π E · Edτ .)
To see how this formual comes about we consider how much work δW is
done when we place an infinitesimal free charge δρf into a potential V . We
have
1
2
Z
δW =
(δρf )V dτ =
where we have used Gauss’ law
1
4πk²o
1
²o ∇
Z
[∇ · (δD)]V dτ
· D = 4πkρf . Using
∇ · (δDV ) = [∇ · (δD)]V + δD · ∇V
= [∇ · (δD)]V − δD · E
we have
Z
∇ · (δDV )dτ
Z
=
I
=
[∇ · (δD)]V dτ −
Z
δD · Edτ
δDV dA
= 0
because D and V are zero on the boundary surface. Thus
δW =
2
1
4πk²o
Z
δD · Edτ
)
Now δ(E
δ(E 2 ) = 2E·δE. Writing δ(D·E
= ²2E·δE = 2E·δD
δE = 2E so that
R
1
1 R
we have δW = 8πk²o δ(D·E) dτ to give W = 4 8πk²o D·Edτ as in equation
(6.29).
6.3. FURTHER DIMENNSIONAL OF ELECTROSTATICS
6.3.3
91
Potential of a Polarized Dielectric
In chapter 4 we found that the dipole term in the general multipole expression of the potential was
V =k
p · R̂
R2
where p is the dipole moment which we defined as p ≡
Z
V =k
R
Pdτ and
P · R̂
R2
(6.30)
and using ∇( R1 ) this becomes
Z
V =k
1
P · ∇( )dτ = k
R
to give
Z
V =k
Z
1
∇ · ( P)dτ − k
R
1
P · da − k
R
Z
Z
1
(∇ · P)dτ
R
1
(∇ · P)dτ
R
(6.31)
R
Now recall that V = k Rq for a point charge and V = k ρdτ
R for a volume
R
for
a
surface
charge.
Thus
the
first
term in (6.31)
charge and V = k σda
R
looks like the potential of a surgace charge σb ≡ P · n, so that P · da = σb da
and the second term looks like the potential of a volume charge ρb = −∇ · P
(see equation 6.10 ). Thus we have
V =k
R
σb da
R
+k
R
ρb dτ
R
(6.32)
where
σb ≡ P · n̂
(6.33)
and
∇ · P = −ρb .
Equation (6.32) means that the potential (and field) of a polarized object is
the same as that produced by a volume charge density ρb = −∇ · P plus a
surface charge density σb ≡ P · n̂.
92
CHAPTER 6. ELECTRO- AND MAGNETOSTATICS IN MATTER
Chapter 7
ELECTRODYNAMICS
AND
MAGNETODYNAMICS
In electrostatics and magnetostatics we can generally treat the electric and
magnetic fields as seperate entities. However when we come to time-dependent
problems, we need to consider combined electromagnetic fields.
7.0.4
Faradays’s Law of Induction
”The discovery in 1820 that there was a close connection between electricity and magnetism was very exciting until then, the two subjects had been
considered as quite independent. The first discovery was that currents in
wires make magnetic fields; then, in the same year, it was found that wires
carrying currents in a magnetic field have forces on them. One of the excitements whenever there is a mechanical force is the probability of using it
in an engine to do work. Almost immediately after their discovery, people
started to design electric motors using the forces on current-carryin wires”
(quoted from Feynman [?], pg.16-1)
Thus physicists found that current i makes magnetic field B. The question was does B make E? Magnetic field cannot move static charges, but
electric field can. (Recall F = q(E + gv × B) ). That is if B makes E it
will make current i flow. Does B make i? People tried various things to
find this our. For instance they put a large magnet next to a wire (to try to
get current to flow) and also put two close wires parallel but no effects were
93
94 CHAPTER 7. ELECTRODYNAMICS AND MAGNETODYNAMICS
observed.
In 1840 Faraday discovered the essential feature that had been missed.
Electric affects only exist when there is something it changing. Thus if a
magnet is moved near a circuit then a current will flow. If one pair of close,
parallel wires has a changing current, then a current is induced in the other.
Thus we say that currents are induced by changing magnetic field.
However it cannot be the magnetic field which directly causes the current
to flow because negative fields produce no forces as stationary charges. The
only thing that can cause a stationary charge to move is an electric field.
Thus the changing magnetic field must somehow be producing a corresponding electric field.
These effects are embodied in Faraday’s law
∇×E+g
∂B
=0
∂dt
where the constant g relates the units (whatever they are chosen to be) of
the two previously unrelated quantities E and B. In integral form Faraday’s
law is
I
E·d` + g
∂φg
=0
∂τ
The emf (electromotive force, which is not actually a force) is defined as
E≡
H
E·d`
(7.1)
so that Faradays law is
E= −g
∂φg
∂t
(7.2)
which says that the emf is a circuit is due to a changing magnetic flux.
The emf, as defined in (7.1) is actually the tangential force (due to dot
product E · d`) per unit charge, integrated over is just a field E.) If you have
an emf E in a circuit then it means that you get current flow according to
Ohm’s law E=iR. Batteries or voltage sources do the same thing.
The emf is also sometimes defined as the work done, per unit charge, by
an energy source in a circuit. Thus
95
H
E≡
W
q
F·d`
q
=
H
E·d`
(7.3)
Actually Faradays complete discovery was that emfs or currents can be
generated in a wire in 3 different ways; 1) by moving the wire near an adjacent
circuit, 2) by changing the current in an adjacent circuit and 3) by moving a
magnet near the wire. All 3 situations agree with the statement that emf or
current is due to a changing magnetic flux linked by the circuit as specified
in Faraday’s law (7.2).
It was Faraday’s observations and experiments that lead him to propose
his law in the form of equation (7.2). This states that the emf (or current) is
proportional to the rate of change of the magnetic flux. One might therefore
think that the constant g is a number to be determined from experiment,
but actually it just depends on the unit system employed. Figure 7.1 gives
a better idea of what is meant by the flux linking the circuit.
The minus sign in Faraday’s law (7.2) is consistent (comes from) Lenz’s
law which states that the direction of the induces currents (and its accompanying magnetic flux) is always such as to oppose the charge of flux through
the circuit. This is illustrated in Figure 7.2. Note that the direction of
the current flow, specified by Lenz’s law, is really nothing more then an
application of F = gqv × B.
Let us illustrate how Lenz’s law works for the current induced in a long,
straight wire segment of a circuit loop in the following example.
Example 7.1.1. Let B0 be the magnetic field inside a current
loop which is produced by an indeuced current of the current
loop within an external magnetic field B as shown in Figure 7.2.
Show that B0 and B have opposite signs if the area increases and
have like signs if the area decreases.
Solution In example 5.3.1 we found that magnetic field due to
a current in a long, straight wire as
B0 =
2km i
d
where d is the distance from the wire. Using Ohm’s law
E= iR
96 CHAPTER 7. ELECTRODYNAMICS AND MAGNETODYNAMICS
we have
2km E
−2km ∂φB
=
g
d R
dR
∂t
B0 =
where we have used (7.2). Writing φB = Ba where a is the area
of the loop inside the external flux B, we have
B0 = −
2km
da
gB
dR
dt
da
where da
dt is the rate of change of the area. Clearly if dt is positive
(area increasing) then B 0 ∝ −B meaning that B0 and B are in
the same direction (B 0 ∝ +B).
Faradays law (7.2) can also be written
E=
H
∂
E·d` = −g ∂t
R
B·da
(7.3)
which clearly shows that flux and be changed by changing B or by changing
the shape or position or orientation of the circuit.
Equation (7.3) is a far
H
reaching generaliztio of Faraday’s law (7.2). d` needn’t be over a closed
electric circuit but can simply be any closed geometrical path in space. Thus
this form of Faraday’s law (7.3) gives a relation between E and B themselves
and suggests a unification of electric and magnetic phenomena.
7.0.5
Analogy between Faraday field and Magnetostatics
( A good reference for this section is pg. 287-289, including Examples 6 and
7 from Griffiths [2]).
A glance at the Maxwell equations (3.1 to 3.4) reveals that electric field
E can arise from two very different sources. Gauss’s law (3.23) tells us that
E can arise from a charge q. while Faraday’s law (3.25) tells us that E can
arise from a changing magnetic flux. Similaraly Ampèré’s law (3.26) tells us
that magnetic field B can arise from a current density j or from a changing
electric flux.
For the static case, Ampèré’s law tells us that magnetic field only aises
due to a current density
∇ × B = 4πkm j
(7.4)
while it is always true that
∇·B=0
(7.5)
7.1. OHM’S LAW AND ELECTROSTATIC FORCE
97
Faraday’s law (non-static) is written
∇ × E = −g
∂B
∂t
(7.6)
Now if the electric field only arises from the changing magnetic flux and if
no change are present then
∇·E=0
(7.7)
also. Thus we have an exact analogy between the magnetic field due to a
static charge density and the electric field due to ∂B
∂t . The integral forms are
I
and
I
B · d` = 4πkm i
(7.8)
∂φB
∂t
(7.9)
E · d` = −g
Thus ”Faraday-induced electric fields are determined by − ∂B
∂t is exactly the
same way as magnetostatic fields are determined by j.” (Griffith pg.287)
The Biot-Savart law
Z
B = km
j × R̂R2 dτ
(7.10)
0
translates into
g
E=−
4π
Z
× R̂
d g
dτ = [
2
R
dt 4π
∂B
∂t
Z
B × R̂
dτ ]
R2
(7.11)
If there is symmetry in the problem we use the techniques of Ampèrian loops
in (7.9) just as we did for (7.8).
(Students study examples 6 and 7 on pg. 288-289 of Griffiths.)
7.1
Ohm’s Law and Electrostatic Force
Introductory physics textbooks usually write Ohm’s law as
V = iR
(7.12)
98 CHAPTER 7. ELECTRODYNAMICS AND MAGNETODYNAMICS
where V is the potential (or voltage) and i is the current flowing though a
resistor R. However books on electromagnetic theory [10, 2] write Ohm’s
law as
j = σE
(7.13)
where j is the current density, σ is the conductivity and E is the electric
field. Let us therefore discuss Ohm’s law from first principles.
Recall the Lorentz force law
F = q(E = gv × B)
(7.14)
which gives the electromagnetic force on a charge q. Obviously the force per
unit charge defined as f ≡ Fq is
f = E + gv × B
(7.15)
Obviously if v or B are zero then the force per unit charge is nothing more
than the electric f = E, but for a moving charge in a magnetic field we have
the more general relation (7.15).
Experimentally it is found that the current flow in a circuit is proportional to the force per unit charge. If you think about it this makes perfect
sense. Actually it is the current density j which is proportional to f and
the proportionality constant is called the conductivity σ. (The resistivity is
defined as R ≡ σ1 .) Thus Ohm’s law is
j=σf
(7.16)
This represents Ohm’s law is its most general form. Two things are worth
noting. Firstly Ohm’s law is an experimental observation. Secondly Ohm’s
law should be regarded just like Hooke’s law F = −kx for a spring. We all
know that Hooke’s law is only valid (experimentally) for small oscillations
and similarly Ohm’s law is only valid for small electrical forces. Just as
Hooke’s law has non-linear corrections for large displacements, so too does
Ohm’s law have neld linear corrections for large currents. Equations (7.16)
yields j = σ(E + f v × B). In circuits usually v and B are quite small, so
that the approximate version is j ≈ σE which is (7.13).
Imagine we have a cylindrical wire of cross sectioned area A, length `
with an electric field E applied in the longitudinal direction as shown in
figure 7.3.
7.1. OHM’S LAW AND ELECTROSTATIC FORCE
99
AssumeR that there is a voltage V between the two ends of the wire. Using
P
V (p) = − ∞
E · d`, it is evident that for a uniform electric field, E can be
taken outside the integral to give E = V` . The current therfore is
i = jA ≈ σEA = σ
V
A
`
(7.17)
showing that the voltage is proportional to the current V∝ i. The resistance
R is the proportionality constant giving V = iR, as given in equation (7.12).
For our example the resistance is
R=
`
`
≡R
σA
A
(7.18)
which, again if you think about it, makes perfect sense.
Just from looking at units, the power P dissipated across a resistor is
P = V i = i2 R
(7.19)
but the more general case uses the proper definition of power as P ≡ bf F · v
which gives
P =F·v =
Z
ρbf E · vdτ =
Z
E · jdτ = σ
Z
E 2 dτ
(7.20)
(NNN see Example 1, Pg.271 of Griffith)
Ohm’s lawHcan be related to the electromotive force E. We defined this
in
before as E≡ E · d`. The electric field E is the force per unit charge
H
the absence of v or B. Thus a more general definition would be E≡ f · d`
which gives
E≡
H
f · d` =
H
bf j
σ
· d` =
H
i
Aσ d`
=i
H
d`
Aσ
= iR.
(7.21)
Thus the emf also obeys the form of Ohm’s law given in (7.12).
Chapter 11
ELECTROMAGNETIC
WAVES
107
108
CHAPTER 11. ELECTROMAGNETIC WAVES
Chapter 12
SPECIAL RELATIVITY
109
110
CHAPTER 12. SPECIAL RELATIVITY
Bibliography
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[2] D.J. Griffiths, Introduction to Electrodynamics, (Prentice-Hall, Englewood Cliffs, New Jersey, 1989). text book for this course.
QC680.G74
[3] J.T. Cushing, Applied Analytical Mathematics for Physical Scientists,
(Wiley, New York, 1975).
[4] J.B. Fraleigh and R.A. Beauregard, Linear Algebra, (Addison-Wesley,
Reading, Masachusetts, 1987).
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(Prentice-Hall, Englewood Cliffs, New Jersey, 1987). QA303.P99
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vols. 1 and 2,
(Addison-Wesley, Reading, Masachusetts, 1969). QC20.B9
[8] G.B. Arfken and H.J. Weber, Mathematical Methods for Physicists, 4th
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QC631.J3
111
112
BIBLIOGRAPHY
[11] J.B. Marion, Classical Electromagnetic Radiation, (Academic Press,
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[12] G.R. Fowles and G.L. Cassiday, Analytical Mechanics, 5th ed. (Saunders
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QA845 .M38
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QC793.5.Q2522 H34